Termination w.r.t. Q proof of TRS/SK902.07.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:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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(g(x, y), z) → F(y, z)
F(g(x, y), z) → F(x, z)
F(f(x, y), z) → F(x, f(y, z))
F(f(x, y), z) → F(y, z)

The TRS R consists of the following rules:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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(g(x, y), z) → F(y, z)
F(g(x, y), z) → F(x, z)
F(f(x, y), z) → F(x, f(y, z))
F(f(x, y), z) → F(y, z)

The TRS R consists of the following rules:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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(g(x, y), z) → F(y, z)
F(g(x, y), z) → F(x, z)
F(f(x, y), z) → F(x, f(y, z))
F(f(x, y), z) → F(y, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2) = F(x1, x2)
g(x1, x2) = g(x1, x2)
f(x1, x2) = f(x1, x2)
1 = 1
0 = 0
2 = 2
i(x1) = i

Recursive path order with status [2].
Precedence:

F₂ > f₂ > i > g₂
1 > g₂
0 > g₂
2 > g₂

Status:

2: multiset
g₂: [1,2]
f₂: [1,2]
1: multiset
0: multiset
F₂: [1,2]
i: multiset

The following usable rules [14] were oriented:

f(1, g(x, y)) → x
f(x, 0) → x
f(f(x, y), z) → f(x, f(y, z))
f(2, g(x, y)) → y
f(g(x, y), z) → g(f(x, z), f(y, z))
f(0, y) → y
f(i(x), y) → i(x)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(0, y) → y
f(x, 0) → x
f(i(x), y) → i(x)
f(f(x, y), z) → f(x, f(y, z))
f(g(x, y), z) → g(f(x, z), f(y, z))
f(1, g(x, y)) → x
f(2, g(x, y)) → y

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.