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