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:
.(.(x, y), z) → .(x, .(y, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
.(.(x, y), z) → .(x, .(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:
.1(.(x, y), z) → .1(y, z)
.1(.(x, y), z) → .1(x, .(y, z))
The TRS R consists of the following rules:
.(.(x, y), z) → .(x, .(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:
.1(.(x, y), z) → .1(y, z)
.1(.(x, y), z) → .1(x, .(y, z))
The TRS R consists of the following rules:
.(.(x, y), z) → .(x, .(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 oriented strictly and are deleted.
.1(.(x, y), z) → .1(y, z)
.1(.(x, y), z) → .1(x, .(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
.1(x1, x2) = .1(x1, x2)
.(x1, x2) = .(x1, x2)
Lexicographic Path Order [19].
Precedence:
[.^12, .2]
The following usable rules [14] were oriented:
.(.(x, y), z) → .(x, .(y, z))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
.(.(x, y), z) → .(x, .(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.