Termination w.r.t. Q proof of TRS/AG01#3.36.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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(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:

G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
F₁(s₁(x)) -> G₁(f₁(x))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
G₁(s₁(x)) -> MINUS₂(s₁(x), f₁(g₁(x)))
F₁(s₁(x)) -> MINUS₂(s₁(x), g₁(f₁(x)))
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

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:

G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
F₁(s₁(x)) -> G₁(f₁(x))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
G₁(s₁(x)) -> MINUS₂(s₁(x), f₁(g₁(x)))
F₁(s₁(x)) -> MINUS₂(s₁(x), g₁(f₁(x)))
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(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.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS₂(x₁, x₂) ) = max{0, x₂ - 2}

POL( s₁(x₁) ) = x₁ + 3

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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
F₁(s₁(x)) -> G₁(f₁(x))
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(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.

G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
F₁(s₁(x)) -> F₁(x)

The remaining pairs can at least be oriented weakly.

F₁(s₁(x)) -> G₁(f₁(x))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( G₁(x₁) ) = max{0, x₁ - 2}

POL( s₁(x₁) ) = x₁ + 3

POL( F₁(x₁) ) = max{0, x₁ - 2}

POL( g₁(x₁) ) = x₁

POL( f₁(x₁) ) = x₁ + 3

POL( 0 ) = 0

POL( minus₂(x₁, x₂) ) = x₁

The following usable rules [14] were oriented:

minus₂(x, 0) -> x
g₁(0) -> 0
f₁(0) -> s₁(0)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

F₁(s₁(x)) -> G₁(f₁(x))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.