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:
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:
A(a(x1)) → A(b(x1))
A(a(x1)) → B(a(b(x1)))
B(b(b(x1))) → A(x1)
A(a(x1)) → B(x1)
B(b(b(x1))) → A(a(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
↳ NonTerminationProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(x1)) → A(b(x1))
A(a(x1)) → B(a(b(x1)))
B(b(b(x1))) → A(x1)
A(a(x1)) → B(x1)
B(b(b(x1))) → A(a(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.
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(a(x1)) → A(b(x1))
A(a(x1)) → B(a(b(x1)))
B(b(b(x1))) → A(x1)
A(a(x1)) → B(x1)
B(b(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(x1))
a(a(x1)) → b(a(b(x1)))
s = A(b(a(a(b(a(a(x1))))))) evaluates to t =A(b(a(a(b(a(a(b(x1))))))))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [x1 / b(x1)]
Rewriting sequence
A(b(a(a(b(a(a(x1))))))) → A(b(a(a(b(b(a(b(x1))))))))
with rule a(a(x1')) → b(a(b(x1'))) at position [0,0,0,0,0] and matcher [x1' / x1]
A(b(a(a(b(b(a(b(x1)))))))) → A(b(b(a(b(b(b(a(b(x1)))))))))
with rule a(a(x1')) → b(a(b(x1'))) at position [0,0] and matcher [x1' / b(b(a(b(x1))))]
A(b(b(a(b(b(b(a(b(x1))))))))) → A(b(b(a(a(a(a(b(x1))))))))
with rule b(b(b(x1'))) → a(a(x1')) at position [0,0,0,0] and matcher [x1' / a(b(x1))]
A(b(b(a(a(a(a(b(x1)))))))) → A(b(b(b(a(b(a(a(b(x1)))))))))
with rule a(a(x1')) → b(a(b(x1'))) at position [0,0,0] and matcher [x1' / a(a(b(x1)))]
A(b(b(b(a(b(a(a(b(x1))))))))) → A(a(a(a(b(a(a(b(x1))))))))
with rule b(b(b(x1'))) → a(a(x1')) at position [0] and matcher [x1' / a(b(a(a(b(x1)))))]
A(a(a(a(b(a(a(b(x1)))))))) → A(b(a(a(b(a(a(b(x1))))))))
with rule A(a(x1)) → A(b(x1))
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(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.