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:

g(x, y) → x
g(x, y) → y
f(s(x), y, y) → f(y, x, s(x))

Q is empty.


QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

g(x, y) → x
g(x, y) → y
f(s(x), y, y) → f(y, x, s(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:

F(s(x), y, y) → F(y, x, s(x))

The TRS R consists of the following rules:

g(x, y) → x
g(x, y) → y
f(s(x), y, y) → f(y, x, s(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:

F(s(x), y, y) → F(y, x, s(x))

The TRS R consists of the following rules:

g(x, y) → x
g(x, y) → y
f(s(x), y, y) → f(y, x, s(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.


F(s(x), y, y) → F(y, x, s(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x1, x2)
s(x1)  =  s(x1)

Recursive path order with status [2].
Precedence:
F2 > s1

Status:
s1: multiset
F2: 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:

g(x, y) → x
g(x, y) → y
f(s(x), y, y) → f(y, x, s(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.