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:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is

g(cons(s(X), Y)) → s(X)
g(cons(0, Y)) → g(Y)

The TRS R 2 is

f(s(X)) → f(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The signature Sigma is {h, f}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

H(cons(X, Y)) → G(cons(X, Y))
G(cons(0, Y)) → G(Y)
H(cons(X, Y)) → H(g(cons(X, Y)))
F(s(X)) → F(X)

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

H(cons(X, Y)) → G(cons(X, Y))
G(cons(0, Y)) → G(Y)
H(cons(X, Y)) → H(g(cons(X, Y)))
F(s(X)) → F(X)

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

H(cons(X, Y)) → G(cons(X, Y))
G(cons(0, Y)) → G(Y)
F(s(X)) → F(X)
H(cons(X, Y)) → H(g(cons(X, Y)))

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

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

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

G(cons(0, Y)) → G(Y)

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

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.

G(cons(0, Y)) → G(Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1) = G(x1)
cons(x1, x2) = cons(x1, x2)
0 = 0

Recursive path order with status [2].
Quasi-Precedence:

cons₂ > [G₁, 0]

Status:

G₁: multiset
0: multiset
cons₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

F(s(X)) → F(X)

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

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)) → F(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1) = F(x1)
s(x1) = s(x1)

Recursive path order with status [2].
Quasi-Precedence:

[F₁, s_1]

Status:

s₁: multiset
F₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

The set Q consists of the following terms:

f(s(x0))
g(cons(0, x0))
g(cons(s(x0), x1))
h(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.