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:
.2(.2(x, y), z) -> .2(x, .2(y, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
.2(.2(x, y), z) -> .2(x, .2(y, z))
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:
.12(.2(x, y), z) -> .12(y, z)
.12(.2(x, y), z) -> .12(x, .2(y, z))
The TRS R consists of the following rules:
.2(.2(x, y), z) -> .2(x, .2(y, z))
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:
.12(.2(x, y), z) -> .12(y, z)
.12(.2(x, y), z) -> .12(x, .2(y, z))
The TRS R consists of the following rules:
.2(.2(x, y), z) -> .2(x, .2(y, z))
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 strictly oriented and are deleted.
.12(.2(x, y), z) -> .12(y, z)
.12(.2(x, y), z) -> .12(x, .2(y, z))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
.12(x1, x2) = .11(x1)
.2(x1, x2) = .2(x1, x2)
Lexicographic Path Order [19].
Precedence:
trivial
The following usable rules [14] were oriented:
.2(.2(x, y), z) -> .2(x, .2(y, z))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
.2(.2(x, y), z) -> .2(x, .2(y, z))
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.