Termination w.r.t. Q proof of TRS/nontermin/Rubio-inntest830.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:

f₁(s₁(X)) -> f₁(X)
g₁(cons₂(0, Y)) -> g₁(Y)
g₁(cons₂(s₁(X), Y)) -> s₁(X)
h₁(cons₂(X, Y)) -> h₁(g₁(cons₂(X, Y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(s₁(X)) -> f₁(X)
g₁(cons₂(0, Y)) -> g₁(Y)
g₁(cons₂(s₁(X), Y)) -> s₁(X)
h₁(cons₂(X, Y)) -> h₁(g₁(cons₂(X, 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:

H₁(cons₂(X, Y)) -> G₁(cons₂(X, Y))
G₁(cons₂(0, Y)) -> G₁(Y)
H₁(cons₂(X, Y)) -> H₁(g₁(cons₂(X, Y)))
F₁(s₁(X)) -> F₁(X)

The TRS R consists of the following rules:

f₁(s₁(X)) -> f₁(X)
g₁(cons₂(0, Y)) -> g₁(Y)
g₁(cons₂(s₁(X), Y)) -> s₁(X)
h₁(cons₂(X, Y)) -> h₁(g₁(cons₂(X, Y)))

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:

H₁(cons₂(X, Y)) -> G₁(cons₂(X, Y))
G₁(cons₂(0, Y)) -> G₁(Y)
H₁(cons₂(X, Y)) -> H₁(g₁(cons₂(X, Y)))
F₁(s₁(X)) -> F₁(X)

The TRS R consists of the following rules:

f₁(s₁(X)) -> f₁(X)
g₁(cons₂(0, Y)) -> g₁(Y)
g₁(cons₂(s₁(X), Y)) -> s₁(X)
h₁(cons₂(X, Y)) -> h₁(g₁(cons₂(X, Y)))

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:

G₁(cons₂(0, Y)) -> G₁(Y)

The TRS R consists of the following rules:

f₁(s₁(X)) -> f₁(X)
g₁(cons₂(0, Y)) -> g₁(Y)
g₁(cons₂(s₁(X), Y)) -> s₁(X)
h₁(cons₂(X, Y)) -> h₁(g₁(cons₂(X, 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.

G₁(cons₂(0, Y)) -> G₁(Y)

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

POL(0) = 3
POL(G₁(x₁)) = x₁ + x₁²
POL(cons₂(x₁, x₂)) = 3 + 3·x₁·x₂ + 2·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:

f₁(s₁(X)) -> f₁(X)
g₁(cons₂(0, Y)) -> g₁(Y)
g₁(cons₂(s₁(X), Y)) -> s₁(X)
h₁(cons₂(X, Y)) -> h₁(g₁(cons₂(X, 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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

F₁(s₁(X)) -> F₁(X)

The TRS R consists of the following rules:

f₁(s₁(X)) -> f₁(X)
g₁(cons₂(0, Y)) -> g₁(Y)
g₁(cons₂(s₁(X), Y)) -> s₁(X)
h₁(cons₂(X, Y)) -> h₁(g₁(cons₂(X, 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.

F₁(s₁(X)) -> F₁(X)

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

POL(F₁(x₁)) = 3·x₁ + 3·x₁²
POL(s₁(x₁)) = 2 + 3·x₁ + 3·x₁²

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:

f₁(s₁(X)) -> f₁(X)
g₁(cons₂(0, Y)) -> g₁(Y)
g₁(cons₂(s₁(X), Y)) -> s₁(X)
h₁(cons₂(X, Y)) -> h₁(g₁(cons₂(X, 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.