Termination w.r.t. Q proof of TRS/AG01#3.53a.trs

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 [2].
Precedence:

s₁ > F₂

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.