Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)
F(x, y, g(z)) → F(x, y, z)

The TRS R consists of the following rules:

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

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(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)
F(x, y, g(z)) → F(x, y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

F(x, y, g(z)) → F(x, y, z)

The remaining pairs can at least be oriented weakly.

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(1) = 0
POL(F(x₁, x₂, x₃)) = x₃
POL(g(x₁)) = 1 + x₁

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

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

F(0, 1, x) → F(g(x), g(x), x)
F(x, g(y), z) → F(x, y, z)
F(g(x), y, z) → F(x, y, z)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.f: 0
g: x₀
0: 1
1: 0
F: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

F.1-0-1(x, g.0(y), z) → F.1-0-1(x, y, z)
F.0-1-1(g.0(x), y, z) → F.0-1-1(x, y, z)
F.1-0-0(x, g.0(y), z) → F.1-0-0(x, y, z)
F.0-1-0(g.0(x), y, z) → F.0-1-0(x, y, z)
F.1-0-1(g.1(x), y, z) → F.1-0-1(x, y, z)
F.1-1-0(g.1(x), y, z) → F.1-1-0(x, y, z)
F.0-0-1(g.0(x), y, z) → F.0-0-1(x, y, z)
F.1-1-1(g.1(x), y, z) → F.1-1-1(x, y, z)
F.0-0-0(g.0(x), y, z) → F.0-0-0(x, y, z)
F.1-1-1(x, g.1(y), z) → F.1-1-1(x, y, z)
F.1-1-0(x, g.1(y), z) → F.1-1-0(x, y, z)
F.0-1-1(x, g.1(y), z) → F.0-1-1(x, y, z)
F.1-0-0(g.1(x), y, z) → F.1-0-0(x, y, z)
F.0-0-0(x, g.0(y), z) → F.0-0-0(x, y, z)
F.1-0-1(0., 1., x) → F.1-1-1(g.1(x), g.1(x), x)
F.0-0-1(x, g.0(y), z) → F.0-0-1(x, y, z)
F.0-1-0(x, g.1(y), z) → F.0-1-0(x, y, z)
F.1-0-0(0., 1., x) → F.0-0-0(g.0(x), g.0(x), x)

The TRS R consists of the following rules:

f.0-1-1(x, y, g.1(z)) → g.0(f.0-1-1(x, y, z))
f.0-1-0(g.0(x), y, z) → g.0(f.0-1-0(x, y, z))
f.0-0-0(x, g.0(y), z) → g.0(f.0-0-0(x, y, z))
f.0-0-1(g.0(x), y, z) → g.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, g.0(z)) → g.0(f.0-1-0(x, y, z))
f.1-0-1(g.1(x), y, z) → g.0(f.1-0-1(x, y, z))
f.0-1-1(x, g.1(y), z) → g.0(f.0-1-1(x, y, z))
f.1-1-0(g.1(x), y, z) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, g.0(y), z) → g.0(f.1-0-1(x, y, z))
f.0-0-0(x, y, g.0(z)) → g.0(f.0-0-0(x, y, z))
f.1-0-0(0., 1., x) → f.0-0-0(g.0(x), g.0(x), x)
f.0-1-1(g.0(x), y, z) → g.0(f.0-1-1(x, y, z))
f.1-0-0(x, g.0(y), z) → g.0(f.1-0-0(x, y, z))
f.0-0-1(x, g.0(y), z) → g.0(f.0-0-1(x, y, z))
f.1-0-1(0., 1., x) → f.1-1-1(g.1(x), g.1(x), x)
f.1-1-1(x, g.1(y), z) → g.0(f.1-1-1(x, y, z))
f.0-0-0(g.0(x), y, z) → g.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, g.0(z)) → g.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, g.0(z)) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, y, g.1(z)) → g.0(f.1-0-1(x, y, z))
f.0-0-1(x, y, g.1(z)) → g.0(f.0-0-1(x, y, z))
f.1-1-1(g.1(x), y, z) → g.0(f.1-1-1(x, y, z))
f.1-1-1(x, y, g.1(z)) → g.0(f.1-1-1(x, y, z))
f.0-1-0(x, g.1(y), z) → g.0(f.0-1-0(x, y, z))
f.1-1-0(x, g.1(y), z) → g.0(f.1-1-0(x, y, z))
f.1-0-0(g.1(x), y, z) → g.0(f.1-0-0(x, y, z))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

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

F.1-0-1(x, g.0(y), z) → F.1-0-1(x, y, z)
F.0-1-1(g.0(x), y, z) → F.0-1-1(x, y, z)
F.1-0-0(x, g.0(y), z) → F.1-0-0(x, y, z)
F.0-1-0(g.0(x), y, z) → F.0-1-0(x, y, z)
F.1-0-1(g.1(x), y, z) → F.1-0-1(x, y, z)
F.1-1-0(g.1(x), y, z) → F.1-1-0(x, y, z)
F.0-0-1(g.0(x), y, z) → F.0-0-1(x, y, z)
F.1-1-1(g.1(x), y, z) → F.1-1-1(x, y, z)
F.0-0-0(g.0(x), y, z) → F.0-0-0(x, y, z)
F.1-1-1(x, g.1(y), z) → F.1-1-1(x, y, z)
F.1-1-0(x, g.1(y), z) → F.1-1-0(x, y, z)
F.0-1-1(x, g.1(y), z) → F.0-1-1(x, y, z)
F.1-0-0(g.1(x), y, z) → F.1-0-0(x, y, z)
F.0-0-0(x, g.0(y), z) → F.0-0-0(x, y, z)
F.1-0-1(0., 1., x) → F.1-1-1(g.1(x), g.1(x), x)
F.0-0-1(x, g.0(y), z) → F.0-0-1(x, y, z)
F.0-1-0(x, g.1(y), z) → F.0-1-0(x, y, z)
F.1-0-0(0., 1., x) → F.0-0-0(g.0(x), g.0(x), x)

The TRS R consists of the following rules:

f.0-1-1(x, y, g.1(z)) → g.0(f.0-1-1(x, y, z))
f.0-1-0(g.0(x), y, z) → g.0(f.0-1-0(x, y, z))
f.0-0-0(x, g.0(y), z) → g.0(f.0-0-0(x, y, z))
f.0-0-1(g.0(x), y, z) → g.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, g.0(z)) → g.0(f.0-1-0(x, y, z))
f.1-0-1(g.1(x), y, z) → g.0(f.1-0-1(x, y, z))
f.0-1-1(x, g.1(y), z) → g.0(f.0-1-1(x, y, z))
f.1-1-0(g.1(x), y, z) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, g.0(y), z) → g.0(f.1-0-1(x, y, z))
f.0-0-0(x, y, g.0(z)) → g.0(f.0-0-0(x, y, z))
f.1-0-0(0., 1., x) → f.0-0-0(g.0(x), g.0(x), x)
f.0-1-1(g.0(x), y, z) → g.0(f.0-1-1(x, y, z))
f.1-0-0(x, g.0(y), z) → g.0(f.1-0-0(x, y, z))
f.0-0-1(x, g.0(y), z) → g.0(f.0-0-1(x, y, z))
f.1-0-1(0., 1., x) → f.1-1-1(g.1(x), g.1(x), x)
f.1-1-1(x, g.1(y), z) → g.0(f.1-1-1(x, y, z))
f.0-0-0(g.0(x), y, z) → g.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, g.0(z)) → g.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, g.0(z)) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, y, g.1(z)) → g.0(f.1-0-1(x, y, z))
f.0-0-1(x, y, g.1(z)) → g.0(f.0-0-1(x, y, z))
f.1-1-1(g.1(x), y, z) → g.0(f.1-1-1(x, y, z))
f.1-1-1(x, y, g.1(z)) → g.0(f.1-1-1(x, y, z))
f.0-1-0(x, g.1(y), z) → g.0(f.0-1-0(x, y, z))
f.1-1-0(x, g.1(y), z) → g.0(f.1-1-0(x, y, z))
f.1-0-0(g.1(x), y, z) → g.0(f.1-0-0(x, y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 8 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

F.1-1-1(g.1(x), y, z) → F.1-1-1(x, y, z)
F.1-1-1(x, g.1(y), z) → F.1-1-1(x, y, z)

The TRS R consists of the following rules:

f.0-1-1(x, y, g.1(z)) → g.0(f.0-1-1(x, y, z))
f.0-1-0(g.0(x), y, z) → g.0(f.0-1-0(x, y, z))
f.0-0-0(x, g.0(y), z) → g.0(f.0-0-0(x, y, z))
f.0-0-1(g.0(x), y, z) → g.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, g.0(z)) → g.0(f.0-1-0(x, y, z))
f.1-0-1(g.1(x), y, z) → g.0(f.1-0-1(x, y, z))
f.0-1-1(x, g.1(y), z) → g.0(f.0-1-1(x, y, z))
f.1-1-0(g.1(x), y, z) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, g.0(y), z) → g.0(f.1-0-1(x, y, z))
f.0-0-0(x, y, g.0(z)) → g.0(f.0-0-0(x, y, z))
f.1-0-0(0., 1., x) → f.0-0-0(g.0(x), g.0(x), x)
f.0-1-1(g.0(x), y, z) → g.0(f.0-1-1(x, y, z))
f.1-0-0(x, g.0(y), z) → g.0(f.1-0-0(x, y, z))
f.0-0-1(x, g.0(y), z) → g.0(f.0-0-1(x, y, z))
f.1-0-1(0., 1., x) → f.1-1-1(g.1(x), g.1(x), x)
f.1-1-1(x, g.1(y), z) → g.0(f.1-1-1(x, y, z))
f.0-0-0(g.0(x), y, z) → g.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, g.0(z)) → g.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, g.0(z)) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, y, g.1(z)) → g.0(f.1-0-1(x, y, z))
f.0-0-1(x, y, g.1(z)) → g.0(f.0-0-1(x, y, z))
f.1-1-1(g.1(x), y, z) → g.0(f.1-1-1(x, y, z))
f.1-1-1(x, y, g.1(z)) → g.0(f.1-1-1(x, y, z))
f.0-1-0(x, g.1(y), z) → g.0(f.0-1-0(x, y, z))
f.1-1-0(x, g.1(y), z) → g.0(f.1-1-0(x, y, z))
f.1-0-0(g.1(x), y, z) → g.0(f.1-0-0(x, y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F.1-1-1(g.1(x), y, z) → F.1-1-1(x, y, z)
F.1-1-1(x, g.1(y), z) → F.1-1-1(x, y, z)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.1-1-1(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(g.1(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

Q DP problem:
P is empty.
R is empty.
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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

F.1-1-0(g.1(x), y, z) → F.1-1-0(x, y, z)
F.1-1-0(x, g.1(y), z) → F.1-1-0(x, y, z)

The TRS R consists of the following rules:

f.0-1-1(x, y, g.1(z)) → g.0(f.0-1-1(x, y, z))
f.0-1-0(g.0(x), y, z) → g.0(f.0-1-0(x, y, z))
f.0-0-0(x, g.0(y), z) → g.0(f.0-0-0(x, y, z))
f.0-0-1(g.0(x), y, z) → g.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, g.0(z)) → g.0(f.0-1-0(x, y, z))
f.1-0-1(g.1(x), y, z) → g.0(f.1-0-1(x, y, z))
f.0-1-1(x, g.1(y), z) → g.0(f.0-1-1(x, y, z))
f.1-1-0(g.1(x), y, z) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, g.0(y), z) → g.0(f.1-0-1(x, y, z))
f.0-0-0(x, y, g.0(z)) → g.0(f.0-0-0(x, y, z))
f.1-0-0(0., 1., x) → f.0-0-0(g.0(x), g.0(x), x)
f.0-1-1(g.0(x), y, z) → g.0(f.0-1-1(x, y, z))
f.1-0-0(x, g.0(y), z) → g.0(f.1-0-0(x, y, z))
f.0-0-1(x, g.0(y), z) → g.0(f.0-0-1(x, y, z))
f.1-0-1(0., 1., x) → f.1-1-1(g.1(x), g.1(x), x)
f.1-1-1(x, g.1(y), z) → g.0(f.1-1-1(x, y, z))
f.0-0-0(g.0(x), y, z) → g.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, g.0(z)) → g.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, g.0(z)) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, y, g.1(z)) → g.0(f.1-0-1(x, y, z))
f.0-0-1(x, y, g.1(z)) → g.0(f.0-0-1(x, y, z))
f.1-1-1(g.1(x), y, z) → g.0(f.1-1-1(x, y, z))
f.1-1-1(x, y, g.1(z)) → g.0(f.1-1-1(x, y, z))
f.0-1-0(x, g.1(y), z) → g.0(f.0-1-0(x, y, z))
f.1-1-0(x, g.1(y), z) → g.0(f.1-1-0(x, y, z))
f.1-0-0(g.1(x), y, z) → g.0(f.1-0-0(x, y, z))

F.1-1-0(g.1(x), y, z) → F.1-1-0(x, y, z)
F.1-1-0(x, g.1(y), z) → F.1-1-0(x, y, z)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.1-1-0(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(g.1(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

Q DP problem:
P is empty.
R is empty.
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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

F.1-0-1(x, g.0(y), z) → F.1-0-1(x, y, z)
F.1-0-1(g.1(x), y, z) → F.1-0-1(x, y, z)

The TRS R consists of the following rules:

f.0-1-1(x, y, g.1(z)) → g.0(f.0-1-1(x, y, z))
f.0-1-0(g.0(x), y, z) → g.0(f.0-1-0(x, y, z))
f.0-0-0(x, g.0(y), z) → g.0(f.0-0-0(x, y, z))
f.0-0-1(g.0(x), y, z) → g.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, g.0(z)) → g.0(f.0-1-0(x, y, z))
f.1-0-1(g.1(x), y, z) → g.0(f.1-0-1(x, y, z))
f.0-1-1(x, g.1(y), z) → g.0(f.0-1-1(x, y, z))
f.1-1-0(g.1(x), y, z) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, g.0(y), z) → g.0(f.1-0-1(x, y, z))
f.0-0-0(x, y, g.0(z)) → g.0(f.0-0-0(x, y, z))
f.1-0-0(0., 1., x) → f.0-0-0(g.0(x), g.0(x), x)
f.0-1-1(g.0(x), y, z) → g.0(f.0-1-1(x, y, z))
f.1-0-0(x, g.0(y), z) → g.0(f.1-0-0(x, y, z))
f.0-0-1(x, g.0(y), z) → g.0(f.0-0-1(x, y, z))
f.1-0-1(0., 1., x) → f.1-1-1(g.1(x), g.1(x), x)
f.1-1-1(x, g.1(y), z) → g.0(f.1-1-1(x, y, z))
f.0-0-0(g.0(x), y, z) → g.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, g.0(z)) → g.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, g.0(z)) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, y, g.1(z)) → g.0(f.1-0-1(x, y, z))
f.0-0-1(x, y, g.1(z)) → g.0(f.0-0-1(x, y, z))
f.1-1-1(g.1(x), y, z) → g.0(f.1-1-1(x, y, z))
f.1-1-1(x, y, g.1(z)) → g.0(f.1-1-1(x, y, z))
f.0-1-0(x, g.1(y), z) → g.0(f.0-1-0(x, y, z))
f.1-1-0(x, g.1(y), z) → g.0(f.1-1-0(x, y, z))
f.1-0-0(g.1(x), y, z) → g.0(f.1-0-0(x, y, z))

F.1-0-1(x, g.0(y), z) → F.1-0-1(x, y, z)
F.1-0-1(g.1(x), y, z) → F.1-0-1(x, y, z)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.1-0-1(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(g.0(x₁)) = x₁
POL(g.1(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

Q DP problem:
P is empty.
R is empty.
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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

F.0-1-1(g.0(x), y, z) → F.0-1-1(x, y, z)
F.0-1-1(x, g.1(y), z) → F.0-1-1(x, y, z)

The TRS R consists of the following rules:

f.0-1-1(x, y, g.1(z)) → g.0(f.0-1-1(x, y, z))
f.0-1-0(g.0(x), y, z) → g.0(f.0-1-0(x, y, z))
f.0-0-0(x, g.0(y), z) → g.0(f.0-0-0(x, y, z))
f.0-0-1(g.0(x), y, z) → g.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, g.0(z)) → g.0(f.0-1-0(x, y, z))
f.1-0-1(g.1(x), y, z) → g.0(f.1-0-1(x, y, z))
f.0-1-1(x, g.1(y), z) → g.0(f.0-1-1(x, y, z))
f.1-1-0(g.1(x), y, z) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, g.0(y), z) → g.0(f.1-0-1(x, y, z))
f.0-0-0(x, y, g.0(z)) → g.0(f.0-0-0(x, y, z))
f.1-0-0(0., 1., x) → f.0-0-0(g.0(x), g.0(x), x)
f.0-1-1(g.0(x), y, z) → g.0(f.0-1-1(x, y, z))
f.1-0-0(x, g.0(y), z) → g.0(f.1-0-0(x, y, z))
f.0-0-1(x, g.0(y), z) → g.0(f.0-0-1(x, y, z))
f.1-0-1(0., 1., x) → f.1-1-1(g.1(x), g.1(x), x)
f.1-1-1(x, g.1(y), z) → g.0(f.1-1-1(x, y, z))
f.0-0-0(g.0(x), y, z) → g.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, g.0(z)) → g.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, g.0(z)) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, y, g.1(z)) → g.0(f.1-0-1(x, y, z))
f.0-0-1(x, y, g.1(z)) → g.0(f.0-0-1(x, y, z))
f.1-1-1(g.1(x), y, z) → g.0(f.1-1-1(x, y, z))
f.1-1-1(x, y, g.1(z)) → g.0(f.1-1-1(x, y, z))
f.0-1-0(x, g.1(y), z) → g.0(f.0-1-0(x, y, z))
f.1-1-0(x, g.1(y), z) → g.0(f.1-1-0(x, y, z))
f.1-0-0(g.1(x), y, z) → g.0(f.1-0-0(x, y, z))

F.0-1-1(g.0(x), y, z) → F.0-1-1(x, y, z)
F.0-1-1(x, g.1(y), z) → F.0-1-1(x, y, z)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-1-1(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(g.0(x₁)) = x₁
POL(g.1(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

Q DP problem:
P is empty.
R is empty.
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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

F.0-1-0(g.0(x), y, z) → F.0-1-0(x, y, z)
F.0-1-0(x, g.1(y), z) → F.0-1-0(x, y, z)

The TRS R consists of the following rules:

f.0-1-1(x, y, g.1(z)) → g.0(f.0-1-1(x, y, z))
f.0-1-0(g.0(x), y, z) → g.0(f.0-1-0(x, y, z))
f.0-0-0(x, g.0(y), z) → g.0(f.0-0-0(x, y, z))
f.0-0-1(g.0(x), y, z) → g.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, g.0(z)) → g.0(f.0-1-0(x, y, z))
f.1-0-1(g.1(x), y, z) → g.0(f.1-0-1(x, y, z))
f.0-1-1(x, g.1(y), z) → g.0(f.0-1-1(x, y, z))
f.1-1-0(g.1(x), y, z) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, g.0(y), z) → g.0(f.1-0-1(x, y, z))
f.0-0-0(x, y, g.0(z)) → g.0(f.0-0-0(x, y, z))
f.1-0-0(0., 1., x) → f.0-0-0(g.0(x), g.0(x), x)
f.0-1-1(g.0(x), y, z) → g.0(f.0-1-1(x, y, z))
f.1-0-0(x, g.0(y), z) → g.0(f.1-0-0(x, y, z))
f.0-0-1(x, g.0(y), z) → g.0(f.0-0-1(x, y, z))
f.1-0-1(0., 1., x) → f.1-1-1(g.1(x), g.1(x), x)
f.1-1-1(x, g.1(y), z) → g.0(f.1-1-1(x, y, z))
f.0-0-0(g.0(x), y, z) → g.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, g.0(z)) → g.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, g.0(z)) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, y, g.1(z)) → g.0(f.1-0-1(x, y, z))
f.0-0-1(x, y, g.1(z)) → g.0(f.0-0-1(x, y, z))
f.1-1-1(g.1(x), y, z) → g.0(f.1-1-1(x, y, z))
f.1-1-1(x, y, g.1(z)) → g.0(f.1-1-1(x, y, z))
f.0-1-0(x, g.1(y), z) → g.0(f.0-1-0(x, y, z))
f.1-1-0(x, g.1(y), z) → g.0(f.1-1-0(x, y, z))
f.1-0-0(g.1(x), y, z) → g.0(f.1-0-0(x, y, z))

F.0-1-0(g.0(x), y, z) → F.0-1-0(x, y, z)
F.0-1-0(x, g.1(y), z) → F.0-1-0(x, y, z)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-1-0(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(g.0(x₁)) = x₁
POL(g.1(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

Q DP problem:
P is empty.
R is empty.
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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                  ↳ QDP

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

F.0-0-1(g.0(x), y, z) → F.0-0-1(x, y, z)
F.0-0-1(x, g.0(y), z) → F.0-0-1(x, y, z)

The TRS R consists of the following rules:

f.0-1-1(x, y, g.1(z)) → g.0(f.0-1-1(x, y, z))
f.0-1-0(g.0(x), y, z) → g.0(f.0-1-0(x, y, z))
f.0-0-0(x, g.0(y), z) → g.0(f.0-0-0(x, y, z))
f.0-0-1(g.0(x), y, z) → g.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, g.0(z)) → g.0(f.0-1-0(x, y, z))
f.1-0-1(g.1(x), y, z) → g.0(f.1-0-1(x, y, z))
f.0-1-1(x, g.1(y), z) → g.0(f.0-1-1(x, y, z))
f.1-1-0(g.1(x), y, z) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, g.0(y), z) → g.0(f.1-0-1(x, y, z))
f.0-0-0(x, y, g.0(z)) → g.0(f.0-0-0(x, y, z))
f.1-0-0(0., 1., x) → f.0-0-0(g.0(x), g.0(x), x)
f.0-1-1(g.0(x), y, z) → g.0(f.0-1-1(x, y, z))
f.1-0-0(x, g.0(y), z) → g.0(f.1-0-0(x, y, z))
f.0-0-1(x, g.0(y), z) → g.0(f.0-0-1(x, y, z))
f.1-0-1(0., 1., x) → f.1-1-1(g.1(x), g.1(x), x)
f.1-1-1(x, g.1(y), z) → g.0(f.1-1-1(x, y, z))
f.0-0-0(g.0(x), y, z) → g.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, g.0(z)) → g.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, g.0(z)) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, y, g.1(z)) → g.0(f.1-0-1(x, y, z))
f.0-0-1(x, y, g.1(z)) → g.0(f.0-0-1(x, y, z))
f.1-1-1(g.1(x), y, z) → g.0(f.1-1-1(x, y, z))
f.1-1-1(x, y, g.1(z)) → g.0(f.1-1-1(x, y, z))
f.0-1-0(x, g.1(y), z) → g.0(f.0-1-0(x, y, z))
f.1-1-0(x, g.1(y), z) → g.0(f.1-1-0(x, y, z))
f.1-0-0(g.1(x), y, z) → g.0(f.1-0-0(x, y, z))

F.0-0-1(g.0(x), y, z) → F.0-0-1(x, y, z)
F.0-0-1(x, g.0(y), z) → F.0-0-1(x, y, z)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-0-1(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(g.0(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

Q DP problem:
P is empty.
R is empty.
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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                  ↳ QDP

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

F.0-0-0(x, g.0(y), z) → F.0-0-0(x, y, z)
F.0-0-0(g.0(x), y, z) → F.0-0-0(x, y, z)

The TRS R consists of the following rules:

f.0-1-1(x, y, g.1(z)) → g.0(f.0-1-1(x, y, z))
f.0-1-0(g.0(x), y, z) → g.0(f.0-1-0(x, y, z))
f.0-0-0(x, g.0(y), z) → g.0(f.0-0-0(x, y, z))
f.0-0-1(g.0(x), y, z) → g.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, g.0(z)) → g.0(f.0-1-0(x, y, z))
f.1-0-1(g.1(x), y, z) → g.0(f.1-0-1(x, y, z))
f.0-1-1(x, g.1(y), z) → g.0(f.0-1-1(x, y, z))
f.1-1-0(g.1(x), y, z) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, g.0(y), z) → g.0(f.1-0-1(x, y, z))
f.0-0-0(x, y, g.0(z)) → g.0(f.0-0-0(x, y, z))
f.1-0-0(0., 1., x) → f.0-0-0(g.0(x), g.0(x), x)
f.0-1-1(g.0(x), y, z) → g.0(f.0-1-1(x, y, z))
f.1-0-0(x, g.0(y), z) → g.0(f.1-0-0(x, y, z))
f.0-0-1(x, g.0(y), z) → g.0(f.0-0-1(x, y, z))
f.1-0-1(0., 1., x) → f.1-1-1(g.1(x), g.1(x), x)
f.1-1-1(x, g.1(y), z) → g.0(f.1-1-1(x, y, z))
f.0-0-0(g.0(x), y, z) → g.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, g.0(z)) → g.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, g.0(z)) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, y, g.1(z)) → g.0(f.1-0-1(x, y, z))
f.0-0-1(x, y, g.1(z)) → g.0(f.0-0-1(x, y, z))
f.1-1-1(g.1(x), y, z) → g.0(f.1-1-1(x, y, z))
f.1-1-1(x, y, g.1(z)) → g.0(f.1-1-1(x, y, z))
f.0-1-0(x, g.1(y), z) → g.0(f.0-1-0(x, y, z))
f.1-1-0(x, g.1(y), z) → g.0(f.1-1-0(x, y, z))
f.1-0-0(g.1(x), y, z) → g.0(f.1-0-0(x, y, z))

F.0-0-0(x, g.0(y), z) → F.0-0-0(x, y, z)
F.0-0-0(g.0(x), y, z) → F.0-0-0(x, y, z)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-0-0(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(g.0(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

Q DP problem:
P is empty.
R is empty.
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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

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

F.1-0-0(x, g.0(y), z) → F.1-0-0(x, y, z)
F.1-0-0(g.1(x), y, z) → F.1-0-0(x, y, z)

The TRS R consists of the following rules:

f.0-1-1(x, y, g.1(z)) → g.0(f.0-1-1(x, y, z))
f.0-1-0(g.0(x), y, z) → g.0(f.0-1-0(x, y, z))
f.0-0-0(x, g.0(y), z) → g.0(f.0-0-0(x, y, z))
f.0-0-1(g.0(x), y, z) → g.0(f.0-0-1(x, y, z))
f.0-1-0(x, y, g.0(z)) → g.0(f.0-1-0(x, y, z))
f.1-0-1(g.1(x), y, z) → g.0(f.1-0-1(x, y, z))
f.0-1-1(x, g.1(y), z) → g.0(f.0-1-1(x, y, z))
f.1-1-0(g.1(x), y, z) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, g.0(y), z) → g.0(f.1-0-1(x, y, z))
f.0-0-0(x, y, g.0(z)) → g.0(f.0-0-0(x, y, z))
f.1-0-0(0., 1., x) → f.0-0-0(g.0(x), g.0(x), x)
f.0-1-1(g.0(x), y, z) → g.0(f.0-1-1(x, y, z))
f.1-0-0(x, g.0(y), z) → g.0(f.1-0-0(x, y, z))
f.0-0-1(x, g.0(y), z) → g.0(f.0-0-1(x, y, z))
f.1-0-1(0., 1., x) → f.1-1-1(g.1(x), g.1(x), x)
f.1-1-1(x, g.1(y), z) → g.0(f.1-1-1(x, y, z))
f.0-0-0(g.0(x), y, z) → g.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, g.0(z)) → g.0(f.1-0-0(x, y, z))
f.1-1-0(x, y, g.0(z)) → g.0(f.1-1-0(x, y, z))
f.1-0-1(x, y, g.1(z)) → g.0(f.1-0-1(x, y, z))
f.0-0-1(x, y, g.1(z)) → g.0(f.0-0-1(x, y, z))
f.1-1-1(g.1(x), y, z) → g.0(f.1-1-1(x, y, z))
f.1-1-1(x, y, g.1(z)) → g.0(f.1-1-1(x, y, z))
f.0-1-0(x, g.1(y), z) → g.0(f.0-1-0(x, y, z))
f.1-1-0(x, g.1(y), z) → g.0(f.1-1-0(x, y, z))
f.1-0-0(g.1(x), y, z) → g.0(f.1-0-0(x, y, z))

F.1-0-0(x, g.0(y), z) → F.1-0-0(x, y, z)
F.1-0-0(g.1(x), y, z) → F.1-0-0(x, y, z)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.1-0-0(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(g.0(x₁)) = x₁
POL(g.1(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ SemLabProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ UsableRulesReductionPairsProof

                      ↳ QDP

                        ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
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.