MAYBE
Termination w.r.t. Q proof of ../tpdb/SRS/Zantema06/loop2.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:
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(b(x1)))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(b(x1)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(b(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(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:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(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:
B(b(b(x1))) → A(x1)
A(a(x1)) → B(x1)
B(b(b(x1))) → A(a(x1))
A(a(x1)) → B(a(b(x1)))
A(a(x1)) → A(b(x1))
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(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:
B(b(b(x1))) → A(x1)
A(a(x1)) → B(x1)
B(b(b(x1))) → A(a(x1))
A(a(x1)) → B(a(b(x1)))
A(a(x1)) → A(b(x1))
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(x1)) → B(a(b(x1))) at position [0] we obtained the following new rules:
A(a(b(b(x0)))) → B(a(a(a(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(x1))) → A(x1)
A(a(x1)) → B(x1)
B(b(b(x1))) → A(a(x1))
A(a(x1)) → A(b(x1))
A(a(b(b(x0)))) → B(a(a(a(x0))))
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(x1)) → A(b(x1)) at position [0] we obtained the following new rules:
A(a(b(b(x0)))) → A(a(a(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(x1))) → A(x1)
A(a(x1)) → B(x1)
A(a(b(b(x0)))) → A(a(a(x0)))
B(b(b(x1))) → A(a(x1))
A(a(b(b(x0)))) → B(a(a(a(x0))))
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(b(x1)))
B(b(b(x1))) → A(x1)
A(a(x1)) → B(x1)
A(a(b(b(x0)))) → A(a(a(x0)))
B(b(b(x1))) → A(a(x1))
A(a(b(b(x0)))) → B(a(a(a(x0))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(b(x1)))
B(b(b(x1))) → A(x1)
A(a(x1)) → B(x1)
A(a(b(b(x0)))) → A(a(a(x0)))
B(b(b(x1))) → A(a(x1))
A(a(b(b(x0)))) → B(a(a(a(x0))))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
b(b(B(x))) → A(x)
a(A(x)) → B(x)
b(b(a(A(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
b(b(a(A(x)))) → a(a(a(B(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
b(b(B(x))) → A(x)
a(A(x)) → B(x)
b(b(a(A(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
b(b(a(A(x)))) → a(a(a(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:
A1(a(x)) → B1(a(b(x)))
B1(b(a(A(x)))) → A1(a(A(x)))
B1(b(B(x))) → A1(A(x))
B1(b(b(x))) → A1(a(x))
B1(b(b(x))) → A1(x)
B1(b(a(A(x)))) → A1(a(a(B(x))))
A1(a(x)) → B1(x)
A1(a(x)) → A1(b(x))
B1(b(a(A(x)))) → A1(B(x))
B1(b(a(A(x)))) → A1(a(B(x)))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
b(b(B(x))) → A(x)
a(A(x)) → B(x)
b(b(a(A(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
b(b(a(A(x)))) → a(a(a(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(a(b(x)))
B1(b(a(A(x)))) → A1(a(A(x)))
B1(b(B(x))) → A1(A(x))
B1(b(b(x))) → A1(a(x))
B1(b(b(x))) → A1(x)
B1(b(a(A(x)))) → A1(a(a(B(x))))
A1(a(x)) → B1(x)
A1(a(x)) → A1(b(x))
B1(b(a(A(x)))) → A1(B(x))
B1(b(a(A(x)))) → A1(a(B(x)))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
b(b(B(x))) → A(x)
a(A(x)) → B(x)
b(b(a(A(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
b(b(a(A(x)))) → a(a(a(B(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ NonTerminationProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(a(b(x)))
B1(b(a(A(x)))) → A1(a(A(x)))
B1(b(b(x))) → A1(a(x))
B1(b(b(x))) → A1(x)
B1(b(a(A(x)))) → A1(a(a(B(x))))
A1(a(x)) → B1(x)
A1(a(x)) → A1(b(x))
B1(b(a(A(x)))) → A1(a(B(x)))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
b(b(B(x))) → A(x)
a(A(x)) → B(x)
b(b(a(A(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
b(b(a(A(x)))) → a(a(a(B(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
The TRS P consists of the following rules:
A1(a(x)) → B1(a(b(x)))
B1(b(a(A(x)))) → A1(a(A(x)))
B1(b(b(x))) → A1(a(x))
B1(b(b(x))) → A1(x)
B1(b(a(A(x)))) → A1(a(a(B(x))))
A1(a(x)) → B1(x)
A1(a(x)) → A1(b(x))
B1(b(a(A(x)))) → A1(a(B(x)))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
b(b(B(x))) → A(x)
a(A(x)) → B(x)
b(b(a(A(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
b(b(a(A(x)))) → a(a(a(B(x))))
s = A1(b(a(a(b(a(A(x'))))))) evaluates to t =A1(b(a(a(b(a(A(x')))))))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
A1(b(a(a(b(a(A(x'))))))) → A1(b(b(a(b(b(a(A(x'))))))))
with rule a(a(x)) → b(a(b(x))) at position [0,0] and matcher [x / b(a(A(x')))]
A1(b(b(a(b(b(a(A(x')))))))) → A1(b(b(a(a(a(A(x')))))))
with rule b(b(a(A(x'')))) → a(a(A(x''))) at position [0,0,0,0] and matcher [x'' / x']
A1(b(b(a(a(a(A(x'))))))) → A1(b(b(b(a(b(a(A(x'))))))))
with rule a(a(x)) → b(a(b(x))) at position [0,0,0] and matcher [x / a(A(x'))]
A1(b(b(b(a(b(a(A(x')))))))) → A1(a(a(a(b(a(A(x')))))))
with rule b(b(b(x''))) → a(a(x'')) at position [0] and matcher [x'' / a(b(a(A(x'))))]
A1(a(a(a(b(a(A(x'))))))) → A1(b(a(a(b(a(A(x')))))))
with rule A1(a(x)) → A1(b(x))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
b(b(B(x))) → A(x)
a(A(x)) → B(x)
b(b(a(A(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
b(b(a(A(x)))) → a(a(a(B(x))))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
B(b(b(x))) → A(x)
A(a(x)) → B(x)
A(a(b(b(x)))) → A(a(a(x)))
B(b(b(x))) → A(a(x))
A(a(b(b(x)))) → B(a(a(a(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
B(b(b(x))) → A(x)
A(a(x)) → B(x)
A(a(b(b(x)))) → A(a(a(x)))
B(b(b(x))) → A(a(x))
A(a(b(b(x)))) → B(a(a(a(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
b(b(B(x))) → A(x)
a(A(x)) → B(x)
b(b(a(A(x)))) → a(a(A(x)))
b(b(B(x))) → a(A(x))
b(b(a(A(x)))) → a(a(a(B(x))))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
B(b(b(x))) → A(x)
A(a(x)) → B(x)
A(a(b(b(x)))) → A(a(a(x)))
B(b(b(x))) → A(a(x))
A(a(b(b(x)))) → B(a(a(a(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
B(b(b(x))) → A(x)
A(a(x)) → B(x)
A(a(b(b(x)))) → A(a(a(x)))
B(b(b(x))) → A(a(x))
A(a(b(b(x)))) → B(a(a(a(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(b(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(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:
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(x)))
Q is empty.