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(a(b(b(x1)))) → b(b(b(b(b(a(a(a(a(a(x1))))))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(b(b(a(a(a(a(a(x1))))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(x1)))) → b(b(b(b(b(a(a(a(a(a(x1))))))))))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(x)))) → a(a(a(a(a(b(b(b(b(b(x))))))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(a(a(b(b(b(b(b(x))))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(x1)))) → b(b(b(b(b(a(a(a(a(a(x1))))))))))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(x)))) → a(a(a(a(a(b(b(b(b(b(x))))))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(a(a(a(b(b(b(b(b(x))))))))))
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(b(b(a(a(a(a(a(x1))))))))))
The set Q consists of the following terms:
a(a(b(b(x0))))
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(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(x1)
A(a(b(b(x1)))) → A(a(a(a(a(x1)))))
A(a(b(b(x1)))) → A(a(a(x1)))
A(a(b(b(x1)))) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(b(b(a(a(a(a(a(x1))))))))))
The set Q consists of the following terms:
a(a(b(b(x0))))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(x1)
A(a(b(b(x1)))) → A(a(a(a(a(x1)))))
A(a(b(b(x1)))) → A(a(a(x1)))
A(a(b(b(x1)))) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(b(b(a(a(a(a(a(x1))))))))))
The set Q consists of the following terms:
a(a(b(b(x0))))
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(a(a(a(a(x1)))))
A(a(b(b(x1)))) → A(x1)
A(a(b(b(x1)))) → A(a(a(x1)))
A(a(b(b(x1)))) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(b(b(a(a(a(a(a(x1))))))))))
Q is empty.
We have to consider all (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(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(a(a(a(a(x1)))))
A(a(b(b(x1)))) → A(x1)
A(a(b(b(x1)))) → A(a(a(x1)))
A(a(b(b(x1)))) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(a(b(b(x1)))) → b(b(b(b(b(a(a(a(a(a(x1))))))))))
s = A(a(a(a(b(b(x1')))))) evaluates to t =A(a(a(a(b(b(b(a(a(a(a(a(x1'))))))))))))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [x1' / b(a(a(a(a(a(x1'))))))]
Rewriting sequence
A(a(a(a(b(b(x1')))))) → A(a(b(b(b(b(b(a(a(a(a(a(x1'))))))))))))
with rule a(a(b(b(x1'')))) → b(b(b(b(b(a(a(a(a(a(x1'')))))))))) at position [0,0] and matcher [x1'' / x1']
A(a(b(b(b(b(b(a(a(a(a(a(x1')))))))))))) → A(a(a(a(b(b(b(a(a(a(a(a(x1'))))))))))))
with rule A(a(b(b(x1)))) → A(a(a(a(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.