Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
REV(++(x, x)) → REV(x)
The TRS R consists of the following rules:
rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
REV(++(x, x)) → REV(x)
The TRS R consists of the following rules:
rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].
The following pairs can be oriented strictly and are deleted.
REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
REV(++(x, x)) → REV(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
REV(x1) = REV(x1)
++(x1, x2) = ++(x1, x2)
Recursive path order with status [2].
Precedence:
++2 > REV1
Status:
++2: multiset
REV1: multiset
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.