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)
.(x1, x2)  =  .(x1, x2)

Recursive Path Order [2].
Precedence:
[.^11, .2]


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:

.(.(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.