MAYBE
Termination w.r.t. Q proof of ../tpdb/SRS/secret06/jambox/1.srs
Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
The set Q is empty.
We have obtained the following QTRS:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
C(a(a(c(x1)))) → A(a(x1))
B(x1) → A(c(a(x1)))
A(b(a(x1))) → B(x1)
B(x1) → C(a(x1))
A(b(a(x1))) → B(b(x1))
C(a(a(c(x1)))) → A(x1)
C(a(a(c(x1)))) → A(a(a(x1)))
B(x1) → A(x1)
The TRS R consists of the following rules:
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(a(c(x1)))) → A(a(x1))
B(x1) → A(c(a(x1)))
A(b(a(x1))) → B(x1)
B(x1) → C(a(x1))
A(b(a(x1))) → B(b(x1))
C(a(a(c(x1)))) → A(x1)
C(a(a(c(x1)))) → A(a(a(x1)))
B(x1) → A(x1)
The TRS R consists of the following rules:
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(a(c(x1)))) → A(a(a(x1))) at position [0] we obtained the following new rules:
C(a(a(c(b(a(x0)))))) → A(a(b(b(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(a(c(x1)))) → A(a(x1))
C(a(a(c(b(a(x0)))))) → A(a(b(b(x0))))
B(x1) → A(c(a(x1)))
B(x1) → C(a(x1))
A(b(a(x1))) → B(x1)
A(b(a(x1))) → B(b(x1))
C(a(a(c(x1)))) → A(x1)
B(x1) → A(x1)
The TRS R consists of the following rules:
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(a(c(x1)))) → A(a(x1)) at position [0] we obtained the following new rules:
C(a(a(c(b(a(x0)))))) → A(b(b(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(a(c(b(a(x0)))))) → A(b(b(x0)))
B(x1) → A(c(a(x1)))
C(a(a(c(b(a(x0)))))) → A(a(b(b(x0))))
A(b(a(x1))) → B(x1)
B(x1) → C(a(x1))
A(b(a(x1))) → B(b(x1))
C(a(a(c(x1)))) → A(x1)
B(x1) → A(x1)
The TRS R consists of the following rules:
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(x1) → A(c(a(x1))) at position [0] we obtained the following new rules:
B(b(a(x0))) → A(c(b(b(x0))))
B(a(c(x0))) → A(a(a(a(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(a(c(b(a(x0)))))) → A(b(b(x0)))
C(a(a(c(b(a(x0)))))) → A(a(b(b(x0))))
B(a(c(x0))) → A(a(a(a(x0))))
B(b(a(x0))) → A(c(b(b(x0))))
B(x1) → C(a(x1))
A(b(a(x1))) → B(x1)
A(b(a(x1))) → B(b(x1))
C(a(a(c(x1)))) → A(x1)
B(x1) → A(x1)
The TRS R consists of the following rules:
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
C(a(a(c(b(a(x0)))))) → A(b(b(x0)))
C(a(a(c(b(a(x0)))))) → A(a(b(b(x0))))
B(a(c(x0))) → A(a(a(a(x0))))
B(b(a(x0))) → A(c(b(b(x0))))
B(x1) → C(a(x1))
A(b(a(x1))) → B(x1)
A(b(a(x1))) → B(b(x1))
C(a(a(c(x1)))) → A(x1)
B(x1) → A(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
C(a(a(c(b(a(x0)))))) → A(b(b(x0)))
C(a(a(c(b(a(x0)))))) → A(a(b(b(x0))))
B(a(c(x0))) → A(a(a(a(x0))))
B(b(a(x0))) → A(c(b(b(x0))))
B(x1) → C(a(x1))
A(b(a(x1))) → B(x1)
A(b(a(x1))) → B(b(x1))
C(a(a(c(x1)))) → A(x1)
B(x1) → A(x1)
The set Q is empty.
We have obtained the following QTRS:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
The set Q is empty.
We have obtained the following QTRS:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
C(a(a(c(b(a(x)))))) → A(b(b(x)))
C(a(a(c(b(a(x)))))) → A(a(b(b(x))))
B(a(c(x))) → A(a(a(a(x))))
B(b(a(x))) → A(c(b(b(x))))
B(x) → C(a(x))
A(b(a(x))) → B(x)
A(b(a(x))) → B(b(x))
C(a(a(c(x)))) → A(x)
B(x) → A(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
C(a(a(c(b(a(x)))))) → A(b(b(x)))
C(a(a(c(b(a(x)))))) → A(a(b(b(x))))
B(a(c(x))) → A(a(a(a(x))))
B(b(a(x))) → A(c(b(b(x))))
B(x) → C(a(x))
A(b(a(x))) → B(x)
A(b(a(x))) → B(b(x))
C(a(a(c(x)))) → A(x)
B(x) → A(x)
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(x)))) → A1(a(a(x)))
C1(a(a(c(x)))) → A1(x)
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
C1(a(B(x))) → A1(A(x))
B1(x) → A1(c(a(x)))
A1(b(B(x))) → B1(c(A(x)))
A1(b(A(x))) → B1(B(x))
B2(x) → A1(C(x))
A1(b(B(x))) → C1(A(x))
C1(a(B(x))) → A1(a(a(A(x))))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(x)))) → A1(a(x))
A1(b(B(x))) → B1(b(c(A(x))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
A1(b(c(a(a(C(x)))))) → A1(A(x))
B1(x) → C1(a(x))
C1(a(B(x))) → A1(a(A(x)))
B1(x) → A1(x)
A1(b(A(x))) → B2(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(x)))) → A1(a(a(x)))
C1(a(a(c(x)))) → A1(x)
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
C1(a(B(x))) → A1(A(x))
B1(x) → A1(c(a(x)))
A1(b(B(x))) → B1(c(A(x)))
A1(b(A(x))) → B1(B(x))
B2(x) → A1(C(x))
A1(b(B(x))) → C1(A(x))
C1(a(B(x))) → A1(a(a(A(x))))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(x)))) → A1(a(x))
A1(b(B(x))) → B1(b(c(A(x))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
A1(b(c(a(a(C(x)))))) → A1(A(x))
B1(x) → C1(a(x))
C1(a(B(x))) → A1(a(A(x)))
B1(x) → A1(x)
A1(b(A(x))) → B2(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 7 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(x)))) → A1(x)
C1(a(a(c(x)))) → A1(a(a(x)))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
B1(x) → A1(c(a(x)))
A1(b(B(x))) → B1(c(A(x)))
A1(b(A(x))) → B1(B(x))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(x)))) → A1(a(x))
A1(b(B(x))) → B1(b(c(A(x))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
B1(x) → C1(a(x))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(x) → A1(c(a(x))) at position [0] we obtained the following new rules:
B1(a(C(x0))) → A1(A(x0))
B1(b(a(x0))) → A1(c(b(b(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
B1(b(A(x0))) → A1(c(b(B(x0))))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(B(x0)) → A1(a(a(a(A(x0)))))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
B1(b(A(x0))) → A1(c(B(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(C(x0))) → A1(A(x0))
C1(a(a(c(x)))) → A1(a(a(x)))
C1(a(a(c(x)))) → A1(x)
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
B1(b(A(x0))) → A1(c(b(B(x0))))
A1(b(B(x))) → B1(c(A(x)))
B1(B(x0)) → A1(a(a(a(A(x0)))))
B1(a(c(x0))) → A1(a(a(a(x0))))
A1(b(A(x))) → B1(B(x))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
B1(b(A(x0))) → A1(c(B(x0)))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(x)))) → A1(a(x))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
A1(b(B(x))) → B1(b(c(A(x))))
B1(b(a(x0))) → A1(c(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(x) → C1(a(x))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(x)))) → A1(x)
C1(a(a(c(x)))) → A1(a(a(x)))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
A1(b(B(x))) → B1(c(A(x)))
B1(b(A(x0))) → A1(c(b(B(x0))))
A1(b(A(x))) → B1(B(x))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
B1(b(A(x0))) → A1(c(B(x0)))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(x)))) → A1(a(x))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
A1(b(B(x))) → B1(b(c(A(x))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
B1(b(a(x0))) → A1(c(b(b(x0))))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
B1(x) → C1(a(x))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(a(c(x)))) → A1(a(a(x))) at position [0] we obtained the following new rules:
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
C1(a(a(c(b(A(x0)))))) → A1(a(B(x0)))
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
A1(b(B(x))) → B1(c(A(x)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
B1(b(A(x0))) → A1(c(B(x0)))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
C1(a(a(c(x)))) → A1(a(x))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
B1(x) → C1(a(x))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
C1(a(a(c(b(A(x0)))))) → A1(a(B(x0)))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
B1(b(A(x0))) → A1(c(b(B(x0))))
A1(b(A(x))) → B1(B(x))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
A1(b(B(x))) → B1(b(c(A(x))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
B1(b(a(x0))) → A1(c(b(b(x0))))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(A(x0))) → A1(c(B(x0))) at position [0] we obtained the following new rules:
B1(b(A(x0))) → A1(c(a(C(x0))))
B1(b(A(x0))) → A1(c(A(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
A1(b(B(x))) → B1(c(A(x)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
C1(a(a(c(x)))) → A1(a(x))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
B1(b(A(x0))) → A1(c(A(x0)))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(x) → C1(a(x))
B1(b(A(x0))) → A1(c(a(C(x0))))
C1(a(a(c(b(A(x0)))))) → A1(a(B(x0)))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
B1(b(A(x0))) → A1(c(b(B(x0))))
A1(b(A(x))) → B1(B(x))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
A1(b(B(x))) → B1(b(c(A(x))))
B1(b(a(x0))) → A1(c(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
A1(b(B(x))) → B1(c(A(x)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
C1(a(a(c(x)))) → A1(a(x))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
B1(x) → C1(a(x))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
C1(a(a(c(b(A(x0)))))) → A1(a(B(x0)))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
B1(b(A(x0))) → A1(c(b(B(x0))))
A1(b(A(x))) → B1(B(x))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
A1(b(B(x))) → B1(b(c(A(x))))
B1(b(a(x0))) → A1(c(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(a(c(x)))) → A1(a(x)) at position [0] we obtained the following new rules:
C1(a(a(c(b(A(x0)))))) → A1(B(x0))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(a(A(x0)))))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(A(x0))))
C1(a(a(c(b(a(x0)))))) → A1(b(b(x0)))
C1(a(a(c(b(B(x0)))))) → A1(b(b(c(A(x0)))))
C1(a(a(c(b(A(x0)))))) → A1(b(B(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(a(A(x0)))))
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
A1(b(B(x))) → B1(c(A(x)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(x) → C1(a(x))
C1(a(a(c(b(A(x0)))))) → A1(B(x0))
C1(a(a(c(b(A(x0)))))) → A1(a(B(x0)))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
C1(a(a(c(b(a(x0)))))) → A1(b(b(x0)))
B1(b(A(x0))) → A1(c(b(B(x0))))
A1(b(A(x))) → B1(B(x))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(A(x0))))
A1(b(B(x))) → B1(b(c(A(x))))
C1(a(a(c(b(B(x0)))))) → A1(b(b(c(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
B1(b(a(x0))) → A1(c(b(b(x0))))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(A(x0)))))) → A1(b(B(x0)))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(a(c(b(A(x0)))))) → A1(a(B(x0))) at position [0] we obtained the following new rules:
C1(a(a(c(b(A(x0)))))) → A1(a(A(x0)))
C1(a(a(c(b(A(x0)))))) → A1(a(a(C(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(a(A(x0)))))
C1(a(a(c(b(A(x0)))))) → A1(a(a(C(x0))))
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
A1(b(B(x))) → B1(c(A(x)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
B1(x) → C1(a(x))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
C1(a(a(c(b(A(x0)))))) → A1(B(x0))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
B1(b(A(x0))) → A1(c(b(B(x0))))
C1(a(a(c(b(a(x0)))))) → A1(b(b(x0)))
A1(b(A(x))) → B1(B(x))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
A1(b(B(x))) → B1(b(c(A(x))))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(A(x0))))
C1(a(a(c(b(A(x0)))))) → A1(a(A(x0)))
B1(b(a(x0))) → A1(c(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
C1(a(a(c(b(B(x0)))))) → A1(b(b(c(A(x0)))))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
C1(a(a(c(b(A(x0)))))) → A1(b(B(x0)))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(a(A(x0)))))
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
A1(b(B(x))) → B1(c(A(x)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(x) → C1(a(x))
C1(a(a(c(b(A(x0)))))) → A1(B(x0))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
C1(a(a(c(b(a(x0)))))) → A1(b(b(x0)))
B1(b(A(x0))) → A1(c(b(B(x0))))
A1(b(A(x))) → B1(B(x))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(A(x0))))
A1(b(B(x))) → B1(b(c(A(x))))
C1(a(a(c(b(B(x0)))))) → A1(b(b(c(A(x0)))))
B1(b(a(x0))) → A1(c(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(A(x0)))))) → A1(b(B(x0)))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(a(a(c(b(A(x0)))))) → A1(B(x0)) at position [0] we obtained the following new rules:
C1(a(a(c(b(A(x0)))))) → A1(a(C(x0)))
C1(a(a(c(b(A(x0)))))) → A1(A(x0))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(a(A(x0)))))
C1(a(a(c(b(A(x0)))))) → A1(a(C(x0)))
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
A1(b(B(x))) → B1(c(A(x)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
C1(a(a(c(b(A(x0)))))) → A1(A(x0))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
B1(x) → C1(a(x))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
B1(b(A(x0))) → A1(c(b(B(x0))))
C1(a(a(c(b(a(x0)))))) → A1(b(b(x0)))
A1(b(A(x))) → B1(B(x))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
A1(b(B(x))) → B1(b(c(A(x))))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(A(x0))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
B1(b(a(x0))) → A1(c(b(b(x0))))
C1(a(a(c(b(B(x0)))))) → A1(b(b(c(A(x0)))))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
C1(a(a(c(b(A(x0)))))) → A1(b(B(x0)))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(a(A(x0)))))
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
A1(b(B(x))) → B1(c(A(x)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
B1(x) → C1(a(x))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
C1(a(a(c(b(a(x0)))))) → A1(b(b(x0)))
B1(b(A(x0))) → A1(c(b(B(x0))))
A1(b(A(x))) → B1(B(x))
A1(b(c(a(a(C(x)))))) → B1(A(x))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
A1(b(B(x))) → B1(b(c(A(x))))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(A(x0))))
C1(a(a(c(b(B(x0)))))) → A1(b(b(c(A(x0)))))
B1(b(a(x0))) → A1(c(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(A(x0)))))) → A1(b(B(x0)))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following quasi-model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.C: 0
C1: 0
a: 1
B1: 0
b: 1
A: 0
c: 1
A1: 0
B: 1
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
B1.0(x) → C1.1(a.0(x))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
B1.1(b.1(B.0(x0))) → A1.0(c.1(b.1(b.1(c.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.1(B.0(x0))))
A1.1(b.1(B.0(x))) → B1.0(b.1(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.1(a.0(A.0(x0))))))
A1.1(b.0(A.1(x))) → B1.0(B.1(x))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(b.1(b.1(c.0(A.0(x0)))))
A1.1(b.0(A.0(x))) → B1.1(B.0(x))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(b.1(a.0(A.1(x))))
B1.1(b.0(A.0(x0))) → A1.0(c.1(b.1(B.0(x0))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.1(a.0(A.1(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.0(b.1(b.1(c.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(b.1(B.1(x0)))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.0(A.1(x0)))))
B1.1(b.1(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
A1.1(b.1(a.0(x))) → B1.0(b.0(x))
B1.0(x) → A1.0(x)
A1.1(b.1(B.1(x))) → B1.1(b.1(c.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.0(a.1(b.1(b.0(x0))))
A1.1(b.1(a.1(x))) → B1.1(x)
A1.1(b.0(A.0(x))) → B1.0(B.0(x))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.0(a.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.0(b.1(b.0(x0)))
A1.1(b.1(B.1(x))) → B1.0(c.0(A.1(x)))
A1.1(b.1(B.1(x))) → B1.1(c.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(b.1(b.1(a.0(A.0(x0)))))
A1.1(b.1(B.0(x))) → B1.0(c.0(A.0(x)))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
C1.1(a.1(a.1(c.0(x)))) → A1.0(x)
B1.1(a.1(c.0(x0))) → A1.0(a.1(a.1(a.0(x0))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.0(b.1(b.1(c.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(b.1(b.1(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.0(x0))))))
A1.1(b.1(a.0(x))) → B1.0(x)
A1.1(b.1(B.0(x))) → B1.1(c.0(A.0(x)))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(b.0(A.1(x)))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.1(a.0(A.0(x0))))))
B1.1(b.1(a.1(x0))) → A1.0(c.1(b.1(b.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(A.0(x))
A1.1(b.1(a.1(x))) → B1.0(x)
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(b.1(B.0(x0)))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(a.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.1(a.0(A.0(x0)))))
B1.1(b.1(B.1(x0))) → A1.1(c.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
B1.1(x) → A1.1(x)
B1.1(b.1(B.1(x0))) → A1.0(c.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(a.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.1(B.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(a.0(A.0(x)))
B1.1(x) → A1.0(x)
B1.1(a.1(c.1(x0))) → A1.0(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(b.1(b.1(c.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(x)))) → A1.0(x)
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.1(a.0(A.1(x))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.0(x0))))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
A1.1(b.1(a.0(x))) → B1.1(b.0(x))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(b.1(a.0(A.0(x))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.1(a.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(b.1(b.1(x0)))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
B1.0(x) → C1.0(a.0(x))
B1.1(b.0(A.1(x0))) → A1.0(c.1(b.1(B.1(x0))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.0(A.0(x0)))))
B1.1(x) → C1.1(a.1(x))
A1.1(b.1(a.1(x))) → B1.0(b.1(x))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(a.1(b.1(B.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.1(a.0(A.1(x0))))))
B1.1(x) → C1.0(a.1(x))
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.1(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(b.1(b.0(A.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(A.1(x))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.0(a.1(b.1(b.1(c.0(A.0(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(a.0(A.0(x)))
A1.1(b.1(B.1(x))) → B1.0(b.1(c.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(a.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.1(B.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.1(B.1(x0)))
B1.1(b.1(B.0(x0))) → A1.1(c.1(b.1(b.1(c.0(A.0(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(a.0(A.1(x)))
B1.1(a.1(c.0(x0))) → A1.1(a.1(a.1(a.0(x0))))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.1(B.1(x0))))
A1.1(b.1(B.0(x))) → B1.1(b.1(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.1(B.0(x0)))
B1.1(b.1(a.0(x0))) → A1.0(c.1(b.1(b.0(x0))))
A1.1(b.0(A.1(x))) → B1.1(B.1(x))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
b.0(x) → a.1(c.1(a.0(x)))
a.1(b.1(B.0(x))) → b.1(b.1(c.0(A.0(x))))
a.1(b.0(A.0(x))) → b.1(B.0(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
c.1(a.1(a.0(C.1(x)))) → A.1(x)
A.1(x0) → A.0(x0)
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
B.1(x0) → B.0(x0)
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.1(a.0(A.1(x))))
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
c.1(a.1(B.0(x))) → a.1(a.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.1(a.0(x))) → b.1(b.0(x))
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
C.1(x0) → C.0(x0)
c.1(a.1(a.1(c.0(x)))) → a.1(a.1(a.0(x)))
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.0(C.0(x)))) → A.0(x)
c.1(a.1(B.1(x))) → a.1(a.1(a.0(A.1(x))))
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
a.1(b.1(B.1(x))) → b.1(b.1(c.0(A.1(x))))
a.1(b.0(A.1(x))) → b.1(B.1(x))
c.1(x0) → c.0(x0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
B1.0(x) → C1.1(a.0(x))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
B1.1(b.1(B.0(x0))) → A1.0(c.1(b.1(b.1(c.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.1(B.0(x0))))
A1.1(b.1(B.0(x))) → B1.0(b.1(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.1(a.0(A.0(x0))))))
A1.1(b.0(A.1(x))) → B1.0(B.1(x))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(b.1(b.1(c.0(A.0(x0)))))
A1.1(b.0(A.0(x))) → B1.1(B.0(x))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(b.1(a.0(A.1(x))))
B1.1(b.0(A.0(x0))) → A1.0(c.1(b.1(B.0(x0))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.1(a.0(A.1(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.0(b.1(b.1(c.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(b.1(B.1(x0)))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.0(A.1(x0)))))
B1.1(b.1(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
A1.1(b.1(a.0(x))) → B1.0(b.0(x))
B1.0(x) → A1.0(x)
A1.1(b.1(B.1(x))) → B1.1(b.1(c.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.0(a.1(b.1(b.0(x0))))
A1.1(b.1(a.1(x))) → B1.1(x)
A1.1(b.0(A.0(x))) → B1.0(B.0(x))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.0(a.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.0(b.1(b.0(x0)))
A1.1(b.1(B.1(x))) → B1.0(c.0(A.1(x)))
A1.1(b.1(B.1(x))) → B1.1(c.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(b.1(b.1(a.0(A.0(x0)))))
A1.1(b.1(B.0(x))) → B1.0(c.0(A.0(x)))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
C1.1(a.1(a.1(c.0(x)))) → A1.0(x)
B1.1(a.1(c.0(x0))) → A1.0(a.1(a.1(a.0(x0))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.0(b.1(b.1(c.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(b.1(b.1(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.0(x0))))))
A1.1(b.1(a.0(x))) → B1.0(x)
A1.1(b.1(B.0(x))) → B1.1(c.0(A.0(x)))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(b.0(A.1(x)))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.1(a.0(A.0(x0))))))
B1.1(b.1(a.1(x0))) → A1.0(c.1(b.1(b.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(A.0(x))
A1.1(b.1(a.1(x))) → B1.0(x)
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(b.1(B.0(x0)))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(a.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.1(a.0(A.0(x0)))))
B1.1(b.1(B.1(x0))) → A1.1(c.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
B1.1(x) → A1.1(x)
B1.1(b.1(B.1(x0))) → A1.0(c.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(a.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.1(B.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(a.0(A.0(x)))
B1.1(x) → A1.0(x)
B1.1(a.1(c.1(x0))) → A1.0(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(b.1(b.1(c.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(x)))) → A1.0(x)
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.1(a.0(A.1(x))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.0(x0))))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
A1.1(b.1(a.0(x))) → B1.1(b.0(x))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(b.1(a.0(A.0(x))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.1(a.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(b.1(b.1(x0)))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
B1.0(x) → C1.0(a.0(x))
B1.1(b.0(A.1(x0))) → A1.0(c.1(b.1(B.1(x0))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.0(A.0(x0)))))
B1.1(x) → C1.1(a.1(x))
A1.1(b.1(a.1(x))) → B1.0(b.1(x))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(a.1(b.1(B.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.1(a.0(A.1(x0))))))
B1.1(x) → C1.0(a.1(x))
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.1(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(b.1(b.0(A.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(A.1(x))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.0(a.1(b.1(b.1(c.0(A.0(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(a.0(A.0(x)))
A1.1(b.1(B.1(x))) → B1.0(b.1(c.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(a.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.1(B.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.1(B.1(x0)))
B1.1(b.1(B.0(x0))) → A1.1(c.1(b.1(b.1(c.0(A.0(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(a.0(A.1(x)))
B1.1(a.1(c.0(x0))) → A1.1(a.1(a.1(a.0(x0))))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.1(B.1(x0))))
A1.1(b.1(B.0(x))) → B1.1(b.1(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.1(B.0(x0)))
B1.1(b.1(a.0(x0))) → A1.0(c.1(b.1(b.0(x0))))
A1.1(b.0(A.1(x))) → B1.1(B.1(x))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
b.0(x) → a.1(c.1(a.0(x)))
a.1(b.1(B.0(x))) → b.1(b.1(c.0(A.0(x))))
a.1(b.0(A.0(x))) → b.1(B.0(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
c.1(a.1(a.0(C.1(x)))) → A.1(x)
A.1(x0) → A.0(x0)
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
B.1(x0) → B.0(x0)
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.1(a.0(A.1(x))))
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
c.1(a.1(B.0(x))) → a.1(a.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.1(a.0(x))) → b.1(b.0(x))
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
C.1(x0) → C.0(x0)
c.1(a.1(a.1(c.0(x)))) → a.1(a.1(a.0(x)))
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.0(C.0(x)))) → A.0(x)
c.1(a.1(B.1(x))) → a.1(a.1(a.0(A.1(x))))
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
a.1(b.1(B.1(x))) → b.1(b.1(c.0(A.1(x))))
a.1(b.0(A.1(x))) → b.1(B.1(x))
c.1(x0) → c.0(x0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 57 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.1(a.0(A.0(x0)))))
B1.1(b.1(B.1(x0))) → A1.1(c.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
B1.1(x) → A1.1(x)
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(b.1(b.1(c.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.1(B.0(x0))))
A1.1(b.0(A.0(x))) → B1.1(B.0(x))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(b.1(b.1(c.0(A.1(x0)))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.0(x0))))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.1(a.0(A.1(x))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
A1.1(b.1(a.0(x))) → B1.1(b.0(x))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.1(a.0(A.0(x))))
B1.1(b.1(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
A1.1(b.1(B.1(x))) → B1.1(b.1(c.0(A.1(x))))
B1.1(x) → C1.1(a.1(x))
A1.1(b.1(a.1(x))) → B1.1(x)
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.1(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.1(x0))))))
A1.1(b.1(B.1(x))) → B1.1(c.0(A.1(x)))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(a.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.1(B.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.1(B.1(x0)))
B1.1(b.1(B.0(x0))) → A1.1(c.1(b.1(b.1(c.0(A.0(x0))))))
B1.1(a.1(c.0(x0))) → A1.1(a.1(a.1(a.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(a.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.1(B.1(x0))))
A1.1(b.1(B.0(x))) → B1.1(b.1(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.1(B.0(x0)))
A1.1(b.1(B.0(x))) → B1.1(c.0(A.0(x)))
A1.1(b.0(A.1(x))) → B1.1(B.1(x))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
b.0(x) → a.1(c.1(a.0(x)))
a.1(b.1(B.0(x))) → b.1(b.1(c.0(A.0(x))))
a.1(b.0(A.0(x))) → b.1(B.0(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
c.1(a.1(a.0(C.1(x)))) → A.1(x)
A.1(x0) → A.0(x0)
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
B.1(x0) → B.0(x0)
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.1(a.0(A.1(x))))
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
c.1(a.1(B.0(x))) → a.1(a.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.1(a.0(x))) → b.1(b.0(x))
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
C.1(x0) → C.0(x0)
c.1(a.1(a.1(c.0(x)))) → a.1(a.1(a.0(x)))
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.0(C.0(x)))) → A.0(x)
c.1(a.1(B.1(x))) → a.1(a.1(a.0(A.1(x))))
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
a.1(b.1(B.1(x))) → b.1(b.1(c.0(A.1(x))))
a.1(b.0(A.1(x))) → b.1(B.1(x))
c.1(x0) → c.0(x0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used.
Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(a(A(x0)))))
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
A1(b(B(x))) → B1(c(A(x)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(x) → C1(a(x))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
B1(b(A(x0))) → A1(c(b(B(x0))))
C1(a(a(c(b(a(x0)))))) → A1(b(b(x0)))
A1(b(A(x))) → B1(B(x))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
A1(b(B(x))) → B1(b(c(A(x))))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(A(x0))))
C1(a(a(c(b(B(x0)))))) → A1(b(b(c(A(x0)))))
B1(b(a(x0))) → A1(c(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(a(A(x)))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(A(x0)))))) → A1(b(B(x0)))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following quasi-model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.C: 0
C1: 0
a: x0
B1: 0
b: 1
A: 0
c: 1
A1: 0
B: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
B1.1(b.0(A.0(x0))) → A1.0(c.1(b.0(B.0(x0))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(a.1(b.0(B.1(x0))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
A1.1(b.0(A.1(x))) → B1.0(B.1(x))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.0(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(a.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.0(a.1(b.1(b.0(x0))))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.0(b.0(A.0(x)))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.1(b.0(a.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.0(a.0(A.1(x0))))))
B1.1(b.0(A.1(x0))) → A1.0(c.1(b.0(B.1(x0))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(b.0(B.1(x0)))
A1.1(b.0(B.1(x))) → B1.1(b.1(c.0(A.1(x))))
B1.1(a.1(c.0(x0))) → A1.0(a.0(a.0(a.0(x0))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.0(B.1(x0))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.0(B.1(x0)))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.0(x0))))))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.0(a.0(A.0(x0))))))
B1.0(x) → A1.0(x)
A1.1(b.0(B.0(x))) → B1.0(b.1(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.0(A.0(x0)))))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(a.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.0(A.1(x0)))))
A1.1(b.1(a.1(x))) → B1.1(x)
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.0(a.0(A.1(x)))
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.0(B.0(x0))))
A1.1(b.0(A.0(x))) → B1.0(B.0(x))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.0(B.0(x0)))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.0(b.1(b.0(a.0(A.1(x0)))))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.0(b.0(a.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.1(b.1(b.1(c.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.0(a.0(A.0(x0))))))
C1.1(a.1(a.1(c.0(x)))) → A1.0(x)
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(a.1(b.0(B.0(x0))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.0(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.0(b.1(b.0(a.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(b.0(B.0(x0)))
B1.1(b.0(a.0(x0))) → A1.0(c.1(b.1(b.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
B1.1(b.1(a.1(x0))) → A1.0(c.1(b.1(b.1(x0))))
A1.1(b.0(a.0(x))) → B1.0(b.0(x))
A1.1(b.1(a.1(x))) → B1.0(x)
B1.1(b.0(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(a.1(b.1(b.1(x0))))
B1.1(x) → A1.1(x)
A1.1(b.0(B.0(x))) → B1.1(c.0(A.0(x)))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.0(A.0(x0)))))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.0(b.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.0(b.1(b.0(x0)))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.0(b.0(a.0(A.0(x))))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.0(A.1(x0)))))
A1.1(b.0(a.0(x))) → B1.1(b.0(x))
A1.1(b.0(a.0(x))) → B1.0(x)
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.0(a.0(A.0(x)))
A1.1(b.0(B.0(x))) → B1.0(c.0(A.0(x)))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.0(b.1(b.1(c.0(A.1(x0)))))
B1.1(x) → A1.0(x)
A1.1(b.0(B.1(x))) → B1.0(c.0(A.1(x)))
B1.1(a.1(c.1(x0))) → A1.0(a.1(a.1(a.1(x0))))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(x)))) → A1.0(x)
B1.1(b.0(B.0(x0))) → A1.1(c.1(b.1(b.1(c.0(A.0(x0))))))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.0(B.1(x0))))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.0(b.1(b.0(A.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(a.0(A.0(x0))))))
A1.1(b.0(B.1(x))) → B1.0(b.1(c.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(b.1(b.1(x0)))
B1.1(b.0(B.1(x0))) → A1.1(c.1(b.1(b.1(c.0(A.1(x0))))))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
B1.1(b.0(B.1(x0))) → A1.0(c.1(b.1(b.1(c.0(A.1(x0))))))
B1.0(x) → C1.0(a.0(x))
B1.1(x) → C1.1(a.1(x))
A1.1(b.1(a.1(x))) → B1.0(b.1(x))
A1.1(b.0(B.0(x))) → B1.1(b.1(c.0(A.0(x))))
B1.1(x) → C1.0(a.1(x))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.0(a.1(b.1(b.1(c.0(A.1(x0))))))
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.0(a.1(b.1(b.1(c.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.0(b.1(b.1(c.0(A.0(x0)))))
B1.1(b.0(B.0(x0))) → A1.0(c.1(b.1(b.1(c.0(A.0(x0))))))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.0(a.0(A.1(x0))))))
A1.1(b.0(B.1(x))) → B1.1(c.0(A.1(x)))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.1(b.1(b.1(c.0(A.1(x0)))))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.1(b.0(a.0(A.0(x))))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
c.0(a.0(B.1(x))) → a.0(a.0(a.0(A.1(x))))
a.1(b.0(A.0(x))) → b.0(B.0(x))
a.1(b.1(c.0(a.0(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
a.1(b.1(c.0(a.0(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.0(A.1(x))) → b.0(B.1(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
A.1(x0) → A.0(x0)
a.1(b.0(a.0(x))) → b.1(b.0(x))
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
B.1(x0) → B.0(x0)
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
a.1(b.1(c.0(a.0(a.0(C.0(x)))))) → b.1(b.0(a.0(A.0(x))))
C.1(x0) → C.0(x0)
c.0(a.0(a.0(C.0(x)))) → A.0(x)
b.0(x) → a.1(c.0(a.0(x)))
c.1(a.1(a.1(c.0(x)))) → a.0(a.0(a.0(x)))
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
c.0(a.0(B.0(x))) → a.0(a.0(a.0(A.0(x))))
a.1(b.1(c.0(a.0(a.0(C.1(x)))))) → b.1(b.0(a.0(A.1(x))))
a.1(b.0(B.0(x))) → b.1(b.1(c.0(A.0(x))))
c.0(a.0(a.0(C.1(x)))) → A.1(x)
c.1(x0) → c.0(x0)
a.1(b.0(B.1(x))) → b.1(b.1(c.0(A.1(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
B1.1(b.0(A.0(x0))) → A1.0(c.1(b.0(B.0(x0))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(a.1(b.0(B.1(x0))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
A1.1(b.0(A.1(x))) → B1.0(B.1(x))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.0(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(a.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.0(a.1(b.1(b.0(x0))))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.0(b.0(A.0(x)))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.1(b.0(a.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.0(a.0(A.1(x0))))))
B1.1(b.0(A.1(x0))) → A1.0(c.1(b.0(B.1(x0))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(b.0(B.1(x0)))
A1.1(b.0(B.1(x))) → B1.1(b.1(c.0(A.1(x))))
B1.1(a.1(c.0(x0))) → A1.0(a.0(a.0(a.0(x0))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.0(B.1(x0))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.0(B.1(x0)))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.0(x0))))))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.0(a.0(A.0(x0))))))
B1.0(x) → A1.0(x)
A1.1(b.0(B.0(x))) → B1.0(b.1(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.0(A.0(x0)))))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(a.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.0(A.1(x0)))))
A1.1(b.1(a.1(x))) → B1.1(x)
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.0(a.0(A.1(x)))
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.0(B.0(x0))))
A1.1(b.0(A.0(x))) → B1.0(B.0(x))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.0(B.0(x0)))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.0(b.1(b.0(a.0(A.1(x0)))))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.0(b.0(a.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.1(b.1(b.1(c.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.0(a.0(A.0(x0))))))
C1.1(a.1(a.1(c.0(x)))) → A1.0(x)
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(a.1(b.0(B.0(x0))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.0(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.0(b.1(b.0(a.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(b.0(B.0(x0)))
B1.1(b.0(a.0(x0))) → A1.0(c.1(b.1(b.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
B1.1(b.1(a.1(x0))) → A1.0(c.1(b.1(b.1(x0))))
A1.1(b.0(a.0(x))) → B1.0(b.0(x))
A1.1(b.1(a.1(x))) → B1.0(x)
B1.1(b.0(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(a.1(b.1(b.1(x0))))
B1.1(x) → A1.1(x)
A1.1(b.0(B.0(x))) → B1.1(c.0(A.0(x)))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.0(A.0(x0)))))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.0(b.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.0(b.1(b.0(x0)))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.0(b.0(a.0(A.0(x))))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.0(A.1(x0)))))
A1.1(b.0(a.0(x))) → B1.1(b.0(x))
A1.1(b.0(a.0(x))) → B1.0(x)
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.0(a.0(A.0(x)))
A1.1(b.0(B.0(x))) → B1.0(c.0(A.0(x)))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.0(b.1(b.1(c.0(A.1(x0)))))
B1.1(x) → A1.0(x)
A1.1(b.0(B.1(x))) → B1.0(c.0(A.1(x)))
B1.1(a.1(c.1(x0))) → A1.0(a.1(a.1(a.1(x0))))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(x)))) → A1.0(x)
B1.1(b.0(B.0(x0))) → A1.1(c.1(b.1(b.1(c.0(A.0(x0))))))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.0(B.1(x0))))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.0(b.1(b.0(A.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(a.0(A.0(x0))))))
A1.1(b.0(B.1(x))) → B1.0(b.1(c.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(b.1(b.1(x0)))
B1.1(b.0(B.1(x0))) → A1.1(c.1(b.1(b.1(c.0(A.1(x0))))))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
B1.1(b.0(B.1(x0))) → A1.0(c.1(b.1(b.1(c.0(A.1(x0))))))
B1.0(x) → C1.0(a.0(x))
B1.1(x) → C1.1(a.1(x))
A1.1(b.1(a.1(x))) → B1.0(b.1(x))
A1.1(b.0(B.0(x))) → B1.1(b.1(c.0(A.0(x))))
B1.1(x) → C1.0(a.1(x))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.0(a.1(b.1(b.1(c.0(A.1(x0))))))
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.0(a.1(b.1(b.1(c.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.0(b.1(b.1(c.0(A.0(x0)))))
B1.1(b.0(B.0(x0))) → A1.0(c.1(b.1(b.1(c.0(A.0(x0))))))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.0(a.0(A.1(x0))))))
A1.1(b.0(B.1(x))) → B1.1(c.0(A.1(x)))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.1(b.1(b.1(c.0(A.1(x0)))))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.1(b.0(a.0(A.0(x))))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
c.0(a.0(B.1(x))) → a.0(a.0(a.0(A.1(x))))
a.1(b.0(A.0(x))) → b.0(B.0(x))
a.1(b.1(c.0(a.0(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
a.1(b.1(c.0(a.0(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.0(A.1(x))) → b.0(B.1(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
A.1(x0) → A.0(x0)
a.1(b.0(a.0(x))) → b.1(b.0(x))
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
B.1(x0) → B.0(x0)
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
a.1(b.1(c.0(a.0(a.0(C.0(x)))))) → b.1(b.0(a.0(A.0(x))))
C.1(x0) → C.0(x0)
c.0(a.0(a.0(C.0(x)))) → A.0(x)
b.0(x) → a.1(c.0(a.0(x)))
c.1(a.1(a.1(c.0(x)))) → a.0(a.0(a.0(x)))
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
c.0(a.0(B.0(x))) → a.0(a.0(a.0(A.0(x))))
a.1(b.1(c.0(a.0(a.0(C.1(x)))))) → b.1(b.0(a.0(A.1(x))))
a.1(b.0(B.0(x))) → b.1(b.1(c.0(A.0(x))))
c.0(a.0(a.0(C.1(x)))) → A.1(x)
c.1(x0) → c.0(x0)
a.1(b.0(B.1(x))) → b.1(b.1(c.0(A.1(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 54 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(a.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
A1.1(b.0(B.0(x))) → B1.1(b.1(c.0(A.0(x))))
A1.1(b.1(a.1(x))) → B1.1(x)
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.0(B.0(x0))))
B1.1(b.0(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.0(B.0(x0)))
A1.1(b.0(B.0(x))) → B1.1(c.0(A.0(x)))
B1.1(x) → A1.1(x)
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
A1.1(b.0(a.0(x))) → B1.1(b.0(x))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(a.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.1(b.1(b.1(c.0(A.0(x0)))))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.1(b.0(a.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.0(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(a.0(A.1(x0))))))
A1.1(b.0(B.1(x))) → B1.1(b.1(c.0(A.1(x))))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.0(B.1(x0))))
B1.1(b.0(B.0(x0))) → A1.1(c.1(b.1(b.1(c.0(A.0(x0))))))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(a.0(A.0(x0))))))
A1.1(b.0(B.1(x))) → B1.1(c.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.0(B.1(x0))))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(a.0(A.1(x0))))))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.0(B.1(x0)))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.1(x0))))))
B1.1(b.0(B.1(x0))) → A1.1(c.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.1(b.1(b.1(c.0(A.1(x0)))))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.1(b.0(a.0(A.0(x))))
B1.1(x) → C1.1(a.1(x))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
c.0(a.0(B.1(x))) → a.0(a.0(a.0(A.1(x))))
a.1(b.0(A.0(x))) → b.0(B.0(x))
a.1(b.1(c.0(a.0(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
a.1(b.1(c.0(a.0(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.0(A.1(x))) → b.0(B.1(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
A.1(x0) → A.0(x0)
a.1(b.0(a.0(x))) → b.1(b.0(x))
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
B.1(x0) → B.0(x0)
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
a.1(b.1(c.0(a.0(a.0(C.0(x)))))) → b.1(b.0(a.0(A.0(x))))
C.1(x0) → C.0(x0)
c.0(a.0(a.0(C.0(x)))) → A.0(x)
b.0(x) → a.1(c.0(a.0(x)))
c.1(a.1(a.1(c.0(x)))) → a.0(a.0(a.0(x)))
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
c.0(a.0(B.0(x))) → a.0(a.0(a.0(A.0(x))))
a.1(b.1(c.0(a.0(a.0(C.1(x)))))) → b.1(b.0(a.0(A.1(x))))
a.1(b.0(B.0(x))) → b.1(b.1(c.0(A.0(x))))
c.0(a.0(a.0(C.1(x)))) → A.1(x)
c.1(x0) → c.0(x0)
a.1(b.0(B.1(x))) → b.1(b.1(c.0(A.1(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
A.1(x0) → A.0(x0)
B.1(x0) → B.0(x0)
C.1(x0) → C.0(x0)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = x1
POL(A.1(x1)) = 1 + x1
POL(A1.1(x1)) = x1
POL(B.0(x1)) = x1
POL(B.1(x1)) = 1 + x1
POL(B1.1(x1)) = x1
POL(C.0(x1)) = x1
POL(C.1(x1)) = 1 + x1
POL(C1.1(x1)) = x1
POL(a.0(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
POL(c.1(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(a.0(A.0(x0))))))
A1.1(b.0(B.0(x))) → B1.1(b.1(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
A1.1(b.1(a.1(x))) → B1.1(x)
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.0(B.0(x0))))
B1.1(b.0(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.0(B.0(x0)))
B1.1(x) → A1.1(x)
A1.1(b.0(B.0(x))) → B1.1(c.0(A.0(x)))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
A1.1(b.0(a.0(x))) → B1.1(b.0(x))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(a.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.1(b.1(b.1(c.0(A.0(x0)))))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.1(b.0(a.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.0(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
B1.1(b.1(c.0(a.0(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(a.0(A.1(x0))))))
A1.1(b.0(B.1(x))) → B1.1(b.1(c.0(A.1(x))))
B1.1(b.0(B.0(x0))) → A1.1(c.1(b.1(b.1(c.0(A.0(x0))))))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.0(B.1(x0))))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(a.0(A.0(x0))))))
A1.1(b.0(B.1(x))) → B1.1(c.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.0(B.1(x0))))
A1.1(b.1(c.0(a.0(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
B1.1(b.1(c.0(a.0(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.0(a.0(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.0(B.1(x0)))
C1.1(a.1(a.1(c.1(b.0(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
C1.1(a.1(a.1(c.1(b.0(B.0(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.0(x0))))))
B1.1(b.0(B.1(x0))) → A1.1(c.1(b.1(b.1(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.1(a.1(b.1(b.1(c.0(A.1(x0))))))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
C1.1(a.1(a.1(c.1(b.0(B.1(x0)))))) → A1.1(b.1(b.1(c.0(A.1(x0)))))
A1.1(b.1(c.0(a.0(a.0(C.0(x)))))) → B1.1(b.0(a.0(A.0(x))))
B1.1(x) → C1.1(a.1(x))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
c.0(a.0(B.1(x))) → a.0(a.0(a.0(A.1(x))))
a.1(b.0(A.0(x))) → b.0(B.0(x))
a.1(b.1(c.0(a.0(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
a.1(b.1(c.0(a.0(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.0(A.1(x))) → b.0(B.1(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
a.1(b.0(a.0(x))) → b.1(b.0(x))
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
a.1(b.1(c.0(a.0(a.0(C.0(x)))))) → b.1(b.0(a.0(A.0(x))))
c.0(a.0(a.0(C.0(x)))) → A.0(x)
b.0(x) → a.1(c.0(a.0(x)))
c.1(a.1(a.1(c.0(x)))) → a.0(a.0(a.0(x)))
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
c.0(a.0(B.0(x))) → a.0(a.0(a.0(A.0(x))))
a.1(b.1(c.0(a.0(a.0(C.1(x)))))) → b.1(b.0(a.0(A.1(x))))
a.1(b.0(B.0(x))) → b.1(b.1(c.0(A.0(x))))
c.0(a.0(a.0(C.1(x)))) → A.1(x)
c.1(x0) → c.0(x0)
a.1(b.0(B.1(x))) → b.1(b.1(c.0(A.1(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used.
Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(a(A(x0)))))
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
A1(b(B(x))) → B1(c(A(x)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(x) → C1(a(x))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
B1(b(A(x0))) → A1(c(b(B(x0))))
C1(a(a(c(b(a(x0)))))) → A1(b(b(x0)))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(A(x0))))
A1(b(B(x))) → B1(b(c(A(x))))
B1(b(a(x0))) → A1(c(b(b(x0))))
C1(a(a(c(b(B(x0)))))) → A1(b(b(c(A(x0)))))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(A(x0)))))) → A1(b(B(x0)))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following quasi-model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.C: 0
C1: 0
a: 1
B1: 0
b: 1
A: 0
c: x0
A1: 0
B: 1
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
B1.0(x) → C1.1(a.0(x))
A1.1(b.1(a.1(x))) → B1.0(x)
A1.1(b.1(B.0(x))) → B1.0(b.0(c.0(A.0(x))))
B1.1(a.0(c.0(x0))) → A1.1(a.1(a.1(a.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
B1.1(b.1(B.0(x0))) → A1.1(c.1(b.1(b.0(c.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.0(b.1(b.0(c.0(A.0(x0)))))
C1.1(a.1(a.0(c.0(x)))) → A1.0(x)
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(b.1(b.0(c.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(b.1(B.0(x0)))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(a.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.1(a.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.0(A.1(x0)))))
B1.1(x) → A1.1(x)
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(a.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.1(a.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.1(B.0(x0))))
B1.1(b.1(B.1(x0))) → A1.0(c.1(b.1(b.0(c.0(A.1(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(b.1(a.0(A.1(x))))
A1.1(b.1(B.1(x))) → B1.0(b.0(c.0(A.1(x))))
B1.1(b.0(A.0(x0))) → A1.0(c.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(b.1(b.0(c.0(A.0(x0)))))
B1.1(x) → A1.0(x)
B1.1(a.1(c.1(x0))) → A1.0(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(x)))) → A1.0(x)
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(a.1(b.1(b.0(c.0(A.0(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.1(a.0(A.1(x))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.1(a.0(A.1(x0))))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.0(x0))))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(b.1(b.0(A.1(x0))))
A1.1(b.1(a.0(x))) → B1.1(b.0(x))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(b.1(a.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.0(b.1(b.0(c.0(A.1(x0)))))
B1.1(b.1(B.0(x0))) → A1.0(c.1(b.1(b.0(c.0(A.0(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.1(a.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(b.1(B.1(x0)))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(b.1(b.1(x0)))
B1.1(b.1(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
A1.1(b.1(a.0(x))) → B1.0(b.0(x))
B1.0(x) → A1.0(x)
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.0(a.1(b.1(b.0(x0))))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
B1.0(x) → C1.0(a.0(x))
B1.1(b.0(A.1(x0))) → A1.0(c.1(b.1(B.1(x0))))
B1.1(x) → C1.1(a.1(x))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.0(A.0(x0)))))
A1.1(b.1(a.1(x))) → B1.0(b.1(x))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(a.1(b.1(B.1(x0))))
A1.1(b.1(B.0(x))) → B1.1(b.0(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.1(a.0(A.1(x0))))))
B1.1(x) → C1.0(a.1(x))
A1.1(b.1(a.1(x))) → B1.1(x)
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.1(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(b.1(b.0(A.0(x0))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.0(b.1(b.0(x0)))
B1.1(b.1(B.1(x0))) → A1.1(c.1(b.1(b.0(c.0(A.1(x0))))))
A1.1(b.1(B.1(x))) → B1.0(c.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.1(x0))))))
B1.1(a.0(c.0(x0))) → A1.0(a.1(a.1(a.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(b.1(b.1(a.0(A.0(x0)))))
A1.1(b.1(B.0(x))) → B1.0(c.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.0(a.1(b.1(b.0(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.1(B.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.1(B.1(x0)))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(a.1(b.1(b.0(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(b.1(b.1(a.0(A.1(x0)))))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.1(B.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.0(x0))))))
A1.1(b.1(B.1(x))) → B1.1(b.0(c.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.1(B.0(x0)))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.0(a.1(b.1(b.0(c.0(A.0(x0))))))
A1.1(b.1(a.0(x))) → B1.0(x)
B1.1(b.1(a.0(x0))) → A1.0(c.1(b.1(b.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(b.0(A.1(x)))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.1(a.0(A.0(x0))))))
B1.1(b.1(a.1(x0))) → A1.0(c.1(b.1(b.1(x0))))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
a.1(b.1(B.0(x))) → b.1(b.0(c.0(A.0(x))))
b.0(x) → a.1(c.1(a.0(x)))
a.1(b.0(A.0(x))) → b.1(B.0(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
c.1(a.1(a.0(C.1(x)))) → A.1(x)
A.1(x0) → A.0(x0)
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
B.1(x0) → B.0(x0)
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.1(a.0(A.1(x))))
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
c.1(a.1(B.0(x))) → a.1(a.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.1(a.0(x))) → b.1(b.0(x))
a.1(b.1(B.1(x))) → b.1(b.0(c.0(A.1(x))))
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
C.1(x0) → C.0(x0)
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.0(C.0(x)))) → A.0(x)
c.1(a.1(B.1(x))) → a.1(a.1(a.0(A.1(x))))
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
c.1(a.1(a.0(c.0(x)))) → a.1(a.1(a.0(x)))
a.1(b.0(A.1(x))) → b.1(B.1(x))
c.1(x0) → c.0(x0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
B1.0(x) → C1.1(a.0(x))
A1.1(b.1(a.1(x))) → B1.0(x)
A1.1(b.1(B.0(x))) → B1.0(b.0(c.0(A.0(x))))
B1.1(a.0(c.0(x0))) → A1.1(a.1(a.1(a.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
B1.1(b.1(B.0(x0))) → A1.1(c.1(b.1(b.0(c.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.0(b.1(b.0(c.0(A.0(x0)))))
C1.1(a.1(a.0(c.0(x)))) → A1.0(x)
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(b.1(b.0(c.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(b.1(B.0(x0)))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(a.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.1(a.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.0(A.1(x0)))))
B1.1(x) → A1.1(x)
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.0(a.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.1(a.0(A.0(x0))))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.1(B.0(x0))))
B1.1(b.1(B.1(x0))) → A1.0(c.1(b.1(b.0(c.0(A.1(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(b.1(a.0(A.1(x))))
A1.1(b.1(B.1(x))) → B1.0(b.0(c.0(A.1(x))))
B1.1(b.0(A.0(x0))) → A1.0(c.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(b.1(b.0(c.0(A.0(x0)))))
B1.1(x) → A1.0(x)
B1.1(a.1(c.1(x0))) → A1.0(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(x)))) → A1.0(x)
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(a.1(b.1(b.0(c.0(A.0(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.1(a.0(A.1(x))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.1(a.0(A.1(x0))))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.0(x0))))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(b.1(b.0(A.1(x0))))
A1.1(b.1(a.0(x))) → B1.1(b.0(x))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.0(b.1(a.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.0(b.1(b.0(c.0(A.1(x0)))))
B1.1(b.1(B.0(x0))) → A1.0(c.1(b.1(b.0(c.0(A.0(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.1(a.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(b.1(B.1(x0)))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.0(c.1(b.1(b.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.0(b.1(b.1(x0)))
B1.1(b.1(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
A1.1(b.1(a.0(x))) → B1.0(b.0(x))
B1.0(x) → A1.0(x)
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.0(a.1(b.1(b.0(x0))))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
B1.0(x) → C1.0(a.0(x))
B1.1(b.0(A.1(x0))) → A1.0(c.1(b.1(B.1(x0))))
B1.1(x) → C1.1(a.1(x))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.0(A.0(x0)))))
A1.1(b.1(a.1(x))) → B1.0(b.1(x))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.0(a.1(b.1(B.1(x0))))
A1.1(b.1(B.0(x))) → B1.1(b.0(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(a.1(b.1(b.1(a.0(A.1(x0))))))
B1.1(x) → C1.0(a.1(x))
A1.1(b.1(a.1(x))) → B1.1(x)
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.1(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(b.1(b.0(A.0(x0))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.0(b.1(b.0(x0)))
B1.1(b.1(B.1(x0))) → A1.1(c.1(b.1(b.0(c.0(A.1(x0))))))
A1.1(b.1(B.1(x))) → B1.0(c.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.1(x0))))))
B1.1(a.0(c.0(x0))) → A1.0(a.1(a.1(a.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(b.1(b.1(a.0(A.0(x0)))))
A1.1(b.1(B.0(x))) → B1.0(c.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.0(a.1(b.1(b.0(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.1(B.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.0(a.1(b.1(b.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.1(B.1(x0)))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(a.1(b.1(b.0(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.0(b.1(b.1(a.0(A.1(x0)))))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.1(B.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.0(x0))))))
A1.1(b.1(B.1(x))) → B1.1(b.0(c.0(A.1(x))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.1(B.0(x0)))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.0(a.1(b.1(b.0(c.0(A.0(x0))))))
A1.1(b.1(a.0(x))) → B1.0(x)
B1.1(b.1(a.0(x0))) → A1.0(c.1(b.1(b.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.0(b.0(A.1(x)))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.0(c.1(b.1(b.1(a.0(A.0(x0))))))
B1.1(b.1(a.1(x0))) → A1.0(c.1(b.1(b.1(x0))))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
a.1(b.1(B.0(x))) → b.1(b.0(c.0(A.0(x))))
b.0(x) → a.1(c.1(a.0(x)))
a.1(b.0(A.0(x))) → b.1(B.0(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
c.1(a.1(a.0(C.1(x)))) → A.1(x)
A.1(x0) → A.0(x0)
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
B.1(x0) → B.0(x0)
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.1(a.0(A.1(x))))
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
c.1(a.1(B.0(x))) → a.1(a.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.1(a.0(x))) → b.1(b.0(x))
a.1(b.1(B.1(x))) → b.1(b.0(c.0(A.1(x))))
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
C.1(x0) → C.0(x0)
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.0(C.0(x)))) → A.0(x)
c.1(a.1(B.1(x))) → a.1(a.1(a.0(A.1(x))))
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
c.1(a.1(a.0(c.0(x)))) → a.1(a.1(a.0(x)))
a.1(b.0(A.1(x))) → b.1(B.1(x))
c.1(x0) → c.0(x0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 51 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
B1.1(a.0(c.0(x0))) → A1.1(a.1(a.1(a.0(x0))))
A1.1(b.1(B.0(x))) → B1.1(b.0(c.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
B1.1(b.1(B.0(x0))) → A1.1(c.1(b.1(b.0(c.0(A.0(x0))))))
A1.1(b.1(a.1(x))) → B1.1(x)
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.1(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(b.1(b.0(c.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.1(a.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.1(x0))))))
B1.1(x) → A1.1(x)
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.1(B.0(x0))))
B1.1(b.1(B.1(x0))) → A1.1(c.1(b.1(b.0(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.1(B.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.1(B.1(x0)))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(b.1(b.0(c.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(a.1(b.1(b.0(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.1(B.1(x0))))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(a.1(b.1(b.0(c.0(A.0(x0))))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.0(x0))))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.1(a.0(A.1(x))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
A1.1(b.1(a.0(x))) → B1.1(b.0(x))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.0(x0))))))
A1.1(b.1(B.1(x))) → B1.1(b.0(c.0(A.1(x))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.1(a.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.1(B.0(x0)))
B1.1(b.1(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
B1.1(x) → C1.1(a.1(x))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
a.1(b.1(B.0(x))) → b.1(b.0(c.0(A.0(x))))
b.0(x) → a.1(c.1(a.0(x)))
a.1(b.0(A.0(x))) → b.1(B.0(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
c.1(a.1(a.0(C.1(x)))) → A.1(x)
A.1(x0) → A.0(x0)
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
B.1(x0) → B.0(x0)
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.1(a.0(A.1(x))))
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
c.1(a.1(B.0(x))) → a.1(a.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.1(a.0(x))) → b.1(b.0(x))
a.1(b.1(B.1(x))) → b.1(b.0(c.0(A.1(x))))
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
C.1(x0) → C.0(x0)
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.0(C.0(x)))) → A.0(x)
c.1(a.1(B.1(x))) → a.1(a.1(a.0(A.1(x))))
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
c.1(a.1(a.0(c.0(x)))) → a.1(a.1(a.0(x)))
a.1(b.0(A.1(x))) → b.1(B.1(x))
c.1(x0) → c.0(x0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
A.1(x0) → A.0(x0)
B.1(x0) → B.0(x0)
C.1(x0) → C.0(x0)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A.0(x1)) = x1
POL(A.1(x1)) = 1 + x1
POL(A1.1(x1)) = x1
POL(B.0(x1)) = x1
POL(B.1(x1)) = 1 + x1
POL(B1.1(x1)) = x1
POL(C.0(x1)) = x1
POL(C.1(x1)) = 1 + x1
POL(C1.1(x1)) = x1
POL(a.0(x1)) = x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
POL(c.0(x1)) = x1
POL(c.1(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ SemLabProof2
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.0(A.1(x0)))))
A1.1(b.1(B.0(x))) → B1.1(b.0(c.0(A.0(x))))
B1.1(a.0(c.0(x0))) → A1.1(a.1(a.1(a.0(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.0(A.1(x)))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(a.1(b.1(b.0(x0))))
B1.1(b.1(B.0(x0))) → A1.1(c.1(b.1(b.0(c.0(A.0(x0))))))
A1.1(b.1(a.1(x))) → B1.1(x)
B1.1(b.1(a.1(x0))) → A1.1(c.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.1(a.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.0(A.0(x0))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(b.1(b.0(c.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(x)))) → A1.1(x)
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(b.1(b.1(a.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(a.0(x0)))))) → A1.1(b.1(b.0(x0)))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.1(x0))))))
B1.1(x) → A1.1(x)
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.0(A.0(x0)))))
B1.1(b.0(A.0(x0))) → A1.1(c.1(b.1(B.0(x0))))
B1.1(b.1(B.1(x0))) → A1.1(c.1(b.1(b.0(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(b.1(b.0(A.1(x0))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(a.1(b.1(B.0(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.1(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(a.1(b.1(B.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.0(A.0(x)))
C1.1(a.1(a.1(c.1(b.0(A.1(x0)))))) → A1.1(b.1(B.1(x0)))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(b.1(b.0(c.0(A.0(x0)))))
C1.1(a.1(a.1(c.1(b.1(B.1(x0)))))) → A1.1(a.1(b.1(b.0(c.0(A.1(x0))))))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(b.1(b.1(x0)))
B1.1(b.0(A.1(x0))) → A1.1(c.1(b.1(B.1(x0))))
A1.1(b.1(c.1(a.1(a.0(C.1(x)))))) → B1.1(b.1(a.0(A.1(x))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.1(a.0(A.0(x0))))))
B1.1(b.1(c.1(a.1(a.0(C.1(x0)))))) → A1.1(c.1(b.1(b.0(A.1(x0)))))
C1.1(a.1(a.1(c.1(b.1(B.0(x0)))))) → A1.1(a.1(b.1(b.0(c.0(A.0(x0))))))
B1.1(b.1(c.1(a.1(a.0(C.0(x0)))))) → A1.1(c.1(b.1(b.0(A.0(x0)))))
B1.1(a.1(c.1(x0))) → A1.1(a.1(a.1(a.1(x0))))
A1.1(b.1(a.0(x))) → B1.1(b.0(x))
C1.1(a.1(a.1(c.1(b.1(a.1(x0)))))) → A1.1(a.1(b.1(b.1(x0))))
C1.1(a.1(a.1(c.1(b.1(c.1(a.1(a.0(C.0(x0))))))))) → A1.1(a.1(b.1(b.1(a.0(A.0(x0))))))
A1.1(b.1(B.1(x))) → B1.1(b.0(c.0(A.1(x))))
A1.1(b.1(c.1(a.1(a.0(C.0(x)))))) → B1.1(b.1(a.0(A.0(x))))
C1.1(a.1(a.1(c.1(b.0(A.0(x0)))))) → A1.1(b.1(B.0(x0)))
B1.1(b.1(a.0(x0))) → A1.1(c.1(b.1(b.0(x0))))
A1.1(b.1(a.1(x))) → B1.1(b.1(x))
B1.1(x) → C1.1(a.1(x))
The TRS R consists of the following rules:
b.1(x) → a.1(c.1(a.1(x)))
a.1(b.1(B.0(x))) → b.1(b.0(c.0(A.0(x))))
b.0(x) → a.1(c.1(a.0(x)))
a.1(b.0(A.0(x))) → b.1(B.0(x))
B.1(x) → A.1(x)
a.1(b.1(a.1(x))) → b.1(b.1(x))
c.1(a.1(a.0(C.1(x)))) → A.1(x)
a.1(b.0(A.1(x))) → B.1(x)
a.1(b.0(A.0(x))) → B.0(x)
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.1(a.0(A.1(x))))
B.1(x) → a.0(C.1(x))
B.0(x) → a.0(C.0(x))
c.1(a.1(B.0(x))) → a.1(a.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.1(a.0(A.0(x))))
a.1(b.1(c.1(a.1(a.0(C.0(x)))))) → b.1(b.0(A.0(x)))
a.1(b.1(a.0(x))) → b.1(b.0(x))
a.1(b.1(B.1(x))) → b.1(b.0(c.0(A.1(x))))
a.1(b.1(c.1(a.1(a.0(C.1(x)))))) → b.1(b.0(A.1(x)))
b.1(x0) → b.0(x0)
a.1(x0) → a.0(x0)
B.0(x) → A.0(x)
c.1(a.1(a.0(C.0(x)))) → A.0(x)
c.1(a.1(B.1(x))) → a.1(a.1(a.0(A.1(x))))
c.1(a.1(a.1(c.1(x)))) → a.1(a.1(a.1(x)))
c.1(a.1(a.0(c.0(x)))) → a.1(a.1(a.0(x)))
a.1(b.0(A.1(x))) → b.1(B.1(x))
c.1(x0) → c.0(x0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
As can be seen after transforming the QDP problem by semantic labelling [33] and then some rule deleting processors, only certain labelled rules and pairs can be used.
Hence, we only have to consider all unlabelled pairs and rules (without the decreasing rules for quasi-models).
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ SemLabProof
↳ SemLabProof2
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(a(A(x0)))))
C1(a(a(c(b(B(x0)))))) → A1(a(b(b(c(A(x0))))))
C1(a(a(c(x)))) → A1(x)
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(A(x0)))))
A1(b(c(a(a(C(x)))))) → B1(b(a(A(x))))
B1(b(A(x0))) → A1(c(b(B(x0))))
C1(a(a(c(b(a(x0)))))) → A1(b(b(x0)))
B1(a(c(x0))) → A1(a(a(a(x0))))
B1(b(c(a(a(C(x0)))))) → A1(c(b(b(a(A(x0))))))
C1(a(a(c(b(a(x0)))))) → A1(a(b(b(x0))))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(a(A(x0))))))
A1(b(B(x))) → B1(b(c(A(x))))
A1(b(c(a(a(C(x)))))) → B1(b(A(x)))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(b(b(A(x0))))
B1(b(a(x0))) → A1(c(b(b(x0))))
C1(a(a(c(b(B(x0)))))) → A1(b(b(c(A(x0)))))
C1(a(a(c(b(A(x0)))))) → A1(a(b(B(x0))))
A1(b(a(x))) → B1(b(x))
A1(b(a(x))) → B1(x)
C1(a(a(c(b(A(x0)))))) → A1(b(B(x0)))
C1(a(a(c(b(c(a(a(C(x0))))))))) → A1(a(b(b(A(x0)))))
B1(x) → C1(a(x))
B1(b(B(x0))) → A1(c(b(b(c(A(x0))))))
B1(x) → A1(x)
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
a(b(c(a(a(C(x)))))) → b(b(A(x)))
a(b(c(a(a(C(x)))))) → b(b(a(A(x))))
c(a(B(x))) → a(a(a(A(x))))
a(b(B(x))) → b(b(c(A(x))))
B(x) → a(C(x))
a(b(A(x))) → B(x)
a(b(A(x))) → b(B(x))
c(a(a(C(x)))) → A(x)
B(x) → A(x)
The set Q is empty.
We have obtained the following QTRS:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
C(a(a(c(b(a(x)))))) → A(b(b(x)))
C(a(a(c(b(a(x)))))) → A(a(b(b(x))))
B(a(c(x))) → A(a(a(a(x))))
B(b(a(x))) → A(c(b(b(x))))
B(x) → C(a(x))
A(b(a(x))) → B(x)
A(b(a(x))) → B(b(x))
C(a(a(c(x)))) → A(x)
B(x) → A(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
C(a(a(c(b(a(x)))))) → A(b(b(x)))
C(a(a(c(b(a(x)))))) → A(a(b(b(x))))
B(a(c(x))) → A(a(a(a(x))))
B(b(a(x))) → A(c(b(b(x))))
B(x) → C(a(x))
A(b(a(x))) → B(x)
A(b(a(x))) → B(b(x))
C(a(a(c(x)))) → A(x)
B(x) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
c(a(a(c(x1)))) → a(a(a(x1)))
b(x1) → a(c(a(x1)))
a(b(a(x1))) → b(b(x1))
The set Q is empty.
We have obtained the following QTRS:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
c(a(a(c(x)))) → a(a(a(x)))
b(x) → a(c(a(x)))
a(b(a(x))) → b(b(x))
Q is empty.