MAYBE
Termination w.r.t. Q proof of ../tpdb/SRS/Waldmann07b/size-12-alpha-3-num-403.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:
a(x1) → b(x1)
b(b(a(c(x1)))) → c(c(a(a(a(a(x1))))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(a(c(x1)))) → c(c(a(a(a(a(x1))))))
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:
B(b(a(c(x1)))) → A(a(a(x1)))
A(x1) → B(x1)
B(b(a(c(x1)))) → A(a(a(a(x1))))
B(b(a(c(x1)))) → A(x1)
B(b(a(c(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(a(c(x1)))) → c(c(a(a(a(a(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(c(x1)))) → A(a(a(x1)))
A(x1) → B(x1)
B(b(a(c(x1)))) → A(a(a(a(x1))))
B(b(a(c(x1)))) → A(x1)
B(b(a(c(x1)))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(a(c(x1)))) → c(c(a(a(a(a(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
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(a(c(x1)))) → c(c(a(a(a(a(x1))))))
B(b(a(c(x1)))) → A(a(a(x1)))
A(x1) → B(x1)
B(b(a(c(x1)))) → A(a(a(a(x1))))
B(b(a(c(x1)))) → A(x1)
B(b(a(c(x1)))) → A(a(x1))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
b(b(a(c(x1)))) → c(c(a(a(a(a(x1))))))
B(b(a(c(x1)))) → A(a(a(x1)))
A(x1) → B(x1)
B(b(a(c(x1)))) → A(a(a(a(x1))))
B(b(a(c(x1)))) → A(x1)
B(b(a(c(x1)))) → A(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
c(a(b(B(x)))) → a(a(A(x)))
A(x) → B(x)
c(a(b(B(x)))) → a(a(a(A(x))))
c(a(b(B(x)))) → A(x)
c(a(b(B(x)))) → a(A(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
c(a(b(B(x)))) → a(a(A(x)))
A(x) → B(x)
c(a(b(B(x)))) → a(a(a(A(x))))
c(a(b(B(x)))) → A(x)
c(a(b(B(x)))) → a(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:
C(a(b(b(x)))) → A1(c(c(x)))
C(a(b(b(x)))) → A1(a(a(c(c(x)))))
C(a(b(B(x)))) → A1(a(a(A(x))))
C(a(b(B(x)))) → A2(x)
C(a(b(B(x)))) → A1(a(A(x)))
C(a(b(b(x)))) → A1(a(c(c(x))))
C(a(b(b(x)))) → A1(a(a(a(c(c(x))))))
C(a(b(B(x)))) → A1(A(x))
C(a(b(b(x)))) → C(x)
C(a(b(b(x)))) → C(c(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
c(a(b(B(x)))) → a(a(A(x)))
A(x) → B(x)
c(a(b(B(x)))) → a(a(a(A(x))))
c(a(b(B(x)))) → A(x)
c(a(b(B(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(b(b(x)))) → A1(c(c(x)))
C(a(b(b(x)))) → A1(a(a(c(c(x)))))
C(a(b(B(x)))) → A1(a(a(A(x))))
C(a(b(B(x)))) → A2(x)
C(a(b(B(x)))) → A1(a(A(x)))
C(a(b(b(x)))) → A1(a(c(c(x))))
C(a(b(b(x)))) → A1(a(a(a(c(c(x))))))
C(a(b(B(x)))) → A1(A(x))
C(a(b(b(x)))) → C(x)
C(a(b(b(x)))) → C(c(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
c(a(b(B(x)))) → a(a(A(x)))
A(x) → B(x)
c(a(b(B(x)))) → a(a(a(A(x))))
c(a(b(B(x)))) → A(x)
c(a(b(B(x)))) → a(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 8 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(b(b(x)))) → C(c(x))
C(a(b(b(x)))) → C(x)
The TRS R consists of the following rules:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
c(a(b(B(x)))) → a(a(A(x)))
A(x) → B(x)
c(a(b(B(x)))) → a(a(a(A(x))))
c(a(b(B(x)))) → A(x)
c(a(b(B(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(b(b(x)))) → C(c(x)) at position [0] we obtained the following new rules:
C(a(b(b(a(b(B(x0))))))) → C(a(A(x0)))
C(a(b(b(a(b(b(x0))))))) → C(a(a(a(a(c(c(x0)))))))
C(a(b(b(a(b(B(x0))))))) → C(A(x0))
C(a(b(b(a(b(B(x0))))))) → C(a(a(a(A(x0)))))
C(a(b(b(a(b(B(x0))))))) → C(a(a(A(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(b(b(a(b(B(x0))))))) → C(A(x0))
C(a(b(b(a(b(B(x0))))))) → C(a(A(x0)))
C(a(b(b(a(b(b(x0))))))) → C(a(a(a(a(c(c(x0)))))))
C(a(b(b(x)))) → C(x)
C(a(b(b(a(b(B(x0))))))) → C(a(a(A(x0))))
C(a(b(b(a(b(B(x0))))))) → C(a(a(a(A(x0)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
c(a(b(B(x)))) → a(a(A(x)))
A(x) → B(x)
c(a(b(B(x)))) → a(a(a(A(x))))
c(a(b(B(x)))) → A(x)
c(a(b(B(x)))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(b(b(a(b(B(x0))))))) → C(A(x0)) at position [0] we obtained the following new rules:
C(a(b(b(a(b(B(x0))))))) → C(B(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(b(b(a(b(B(x0))))))) → C(a(A(x0)))
C(a(b(b(a(b(b(x0))))))) → C(a(a(a(a(c(c(x0)))))))
C(a(b(b(x)))) → C(x)
C(a(b(b(a(b(B(x0))))))) → C(B(x0))
C(a(b(b(a(b(B(x0))))))) → C(a(a(a(A(x0)))))
C(a(b(b(a(b(B(x0))))))) → C(a(a(A(x0))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
c(a(b(B(x)))) → a(a(A(x)))
A(x) → B(x)
c(a(b(B(x)))) → a(a(a(A(x))))
c(a(b(B(x)))) → A(x)
c(a(b(B(x)))) → a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(b(b(a(b(B(x0))))))) → C(a(A(x0)))
C(a(b(b(a(b(b(x0))))))) → C(a(a(a(a(c(c(x0)))))))
C(a(b(b(x)))) → C(x)
C(a(b(b(a(b(B(x0))))))) → C(a(a(A(x0))))
C(a(b(b(a(b(B(x0))))))) → C(a(a(a(A(x0)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
c(a(b(B(x)))) → a(a(A(x)))
A(x) → B(x)
c(a(b(B(x)))) → a(a(a(A(x))))
c(a(b(B(x)))) → A(x)
c(a(b(B(x)))) → a(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
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
c(a(b(B(x)))) → a(a(A(x)))
A(x) → B(x)
c(a(b(B(x)))) → a(a(a(A(x))))
c(a(b(B(x)))) → A(x)
c(a(b(B(x)))) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(b(a(c(x)))) → c(c(a(a(a(a(x))))))
B(b(a(c(x)))) → A(a(a(x)))
A(x) → B(x)
B(b(a(c(x)))) → A(a(a(a(x))))
B(b(a(c(x)))) → A(x)
B(b(a(c(x)))) → A(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
b(b(a(c(x)))) → c(c(a(a(a(a(x))))))
B(b(a(c(x)))) → A(a(a(x)))
A(x) → B(x)
B(b(a(c(x)))) → A(a(a(a(x))))
B(b(a(c(x)))) → A(x)
B(b(a(c(x)))) → A(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
c(a(b(B(x)))) → a(a(A(x)))
A(x) → B(x)
c(a(b(B(x)))) → a(a(a(A(x))))
c(a(b(B(x)))) → A(x)
c(a(b(B(x)))) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(b(a(c(x)))) → c(c(a(a(a(a(x))))))
B(b(a(c(x)))) → A(a(a(x)))
A(x) → B(x)
B(b(a(c(x)))) → A(a(a(a(x))))
B(b(a(c(x)))) → A(x)
B(b(a(c(x)))) → A(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
b(b(a(c(x)))) → c(c(a(a(a(a(x))))))
B(b(a(c(x)))) → A(a(a(x)))
A(x) → B(x)
B(b(a(c(x)))) → A(a(a(a(x))))
B(b(a(c(x)))) → A(x)
B(b(a(c(x)))) → A(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
b(b(a(c(x1)))) → c(c(a(a(a(a(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
b(b(a(c(x1)))) → c(c(a(a(a(a(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
c(a(b(b(x)))) → a(a(a(a(c(c(x))))))
Q is empty.