Termination w.r.t. Q proof of TRS/SK904.52.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:

s₁(a) -> a
s₁(s₁(x)) -> x
s₁(f₂(x, y)) -> f₂(s₁(y), s₁(x))
s₁(g₂(x, y)) -> g₂(s₁(x), s₁(y))
f₂(x, a) -> x
f₂(a, y) -> y
f₂(g₂(x, y), g₂(u, v)) -> g₂(f₂(x, u), f₂(y, v))
g₂(a, a) -> a

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

s₁(a) -> a
s₁(s₁(x)) -> x
s₁(f₂(x, y)) -> f₂(s₁(y), s₁(x))
s₁(g₂(x, y)) -> g₂(s₁(x), s₁(y))
f₂(x, a) -> x
f₂(a, y) -> y
f₂(g₂(x, y), g₂(u, v)) -> g₂(f₂(x, u), f₂(y, v))
g₂(a, a) -> a

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:

S₁(g₂(x, y)) -> S₁(y)
S₁(g₂(x, y)) -> S₁(x)
S₁(f₂(x, y)) -> S₁(x)
S₁(g₂(x, y)) -> G₂(s₁(x), s₁(y))
F₂(g₂(x, y), g₂(u, v)) -> G₂(f₂(x, u), f₂(y, v))
S₁(f₂(x, y)) -> F₂(s₁(y), s₁(x))
S₁(f₂(x, y)) -> S₁(y)
F₂(g₂(x, y), g₂(u, v)) -> F₂(y, v)
F₂(g₂(x, y), g₂(u, v)) -> F₂(x, u)

The TRS R consists of the following rules:

s₁(a) -> a
s₁(s₁(x)) -> x
s₁(f₂(x, y)) -> f₂(s₁(y), s₁(x))
s₁(g₂(x, y)) -> g₂(s₁(x), s₁(y))
f₂(x, a) -> x
f₂(a, y) -> y
f₂(g₂(x, y), g₂(u, v)) -> g₂(f₂(x, u), f₂(y, v))
g₂(a, a) -> a

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

S₁(g₂(x, y)) -> S₁(y)
S₁(g₂(x, y)) -> S₁(x)
S₁(f₂(x, y)) -> S₁(x)
S₁(g₂(x, y)) -> G₂(s₁(x), s₁(y))
F₂(g₂(x, y), g₂(u, v)) -> G₂(f₂(x, u), f₂(y, v))
S₁(f₂(x, y)) -> F₂(s₁(y), s₁(x))
S₁(f₂(x, y)) -> S₁(y)
F₂(g₂(x, y), g₂(u, v)) -> F₂(y, v)
F₂(g₂(x, y), g₂(u, v)) -> F₂(x, u)

The TRS R consists of the following rules:

s₁(a) -> a
s₁(s₁(x)) -> x
s₁(f₂(x, y)) -> f₂(s₁(y), s₁(x))
s₁(g₂(x, y)) -> g₂(s₁(x), s₁(y))
f₂(x, a) -> x
f₂(a, y) -> y
f₂(g₂(x, y), g₂(u, v)) -> g₂(f₂(x, u), f₂(y, v))
g₂(a, a) -> a

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

F₂(g₂(x, y), g₂(u, v)) -> F₂(x, u)
F₂(g₂(x, y), g₂(u, v)) -> F₂(y, v)

The TRS R consists of the following rules:

s₁(a) -> a
s₁(s₁(x)) -> x
s₁(f₂(x, y)) -> f₂(s₁(y), s₁(x))
s₁(g₂(x, y)) -> g₂(s₁(x), s₁(y))
f₂(x, a) -> x
f₂(a, y) -> y
f₂(g₂(x, y), g₂(u, v)) -> g₂(f₂(x, u), f₂(y, v))
g₂(a, a) -> a

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.

F₂(g₂(x, y), g₂(u, v)) -> F₂(x, u)
F₂(g₂(x, y), g₂(u, v)) -> F₂(y, v)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = x₂
POL(g₂(x₁, x₂)) = 1 + x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

s₁(a) -> a
s₁(s₁(x)) -> x
s₁(f₂(x, y)) -> f₂(s₁(y), s₁(x))
s₁(g₂(x, y)) -> g₂(s₁(x), s₁(y))
f₂(x, a) -> x
f₂(a, y) -> y
f₂(g₂(x, y), g₂(u, v)) -> g₂(f₂(x, u), f₂(y, v))
g₂(a, a) -> a

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

S₁(g₂(x, y)) -> S₁(y)
S₁(g₂(x, y)) -> S₁(x)
S₁(f₂(x, y)) -> S₁(x)
S₁(f₂(x, y)) -> S₁(y)

The TRS R consists of the following rules:

s₁(a) -> a
s₁(s₁(x)) -> x
s₁(f₂(x, y)) -> f₂(s₁(y), s₁(x))
s₁(g₂(x, y)) -> g₂(s₁(x), s₁(y))
f₂(x, a) -> x
f₂(a, y) -> y
f₂(g₂(x, y), g₂(u, v)) -> g₂(f₂(x, u), f₂(y, v))
g₂(a, a) -> a

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.

S₁(g₂(x, y)) -> S₁(y)
S₁(g₂(x, y)) -> S₁(x)
S₁(f₂(x, y)) -> S₁(x)
S₁(f₂(x, y)) -> S₁(y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(S₁(x₁)) = x₁
POL(f₂(x₁, x₂)) = 1 + x₁ + x₂
POL(g₂(x₁, x₂)) = 1 + x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

s₁(a) -> a
s₁(s₁(x)) -> x
s₁(f₂(x, y)) -> f₂(s₁(y), s₁(x))
s₁(g₂(x, y)) -> g₂(s₁(x), s₁(y))
f₂(x, a) -> x
f₂(a, y) -> y
f₂(g₂(x, y), g₂(u, v)) -> g₂(f₂(x, u), f₂(y, v))
g₂(a, a) -> a

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.