Termination w.r.t. Q of the following Term Rewriting System could be disproven:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(a(b(x)))) → b(a(b(a(a(b(x))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(a(b(x)))) → b(a(b(a(a(b(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(a(b(x)))) → b(a(b(a(a(b(x))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(a(x)))) → b(a(a(b(a(b(x))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(a(x)))) → b(a(a(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:
A(b(a(b(x)))) → A(a(b(x)))
A(b(a(b(x)))) → A(b(a(a(b(x)))))
The TRS R consists of the following rules:
a(b(a(b(x)))) → b(a(b(a(a(b(x))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
A(b(a(b(x)))) → A(a(b(x)))
A(b(a(b(x)))) → A(b(a(a(b(x)))))
The TRS R consists of the following rules:
a(b(a(b(x)))) → b(a(b(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:
A(b(a(b(x)))) → A(a(b(x)))
A(b(a(b(x)))) → A(b(a(a(b(x)))))
The TRS R consists of the following rules:
a(b(a(b(x)))) → b(a(b(a(a(b(x))))))
s = A(b(a(a(b(a(b(x'))))))) evaluates to t =A(b(a(a(b(a(b(a(a(b(x'))))))))))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [x' / a(a(b(x')))]
Rewriting sequence
A(b(a(a(b(a(b(x'))))))) → A(b(a(b(a(b(a(a(b(x')))))))))
with rule a(b(a(b(x'')))) → b(a(b(a(a(b(x'')))))) at position [0,0,0] and matcher [x'' / x']
A(b(a(b(a(b(a(a(b(x'))))))))) → A(b(a(a(b(a(b(a(a(b(x'))))))))))
with rule A(b(a(b(x)))) → A(b(a(a(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.