Termination w.r.t. Q proof of /tmp/tmpQiFTfb/jwtpa1.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(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

The set Q consists of the following terms:

f(f(a, x0), x1)

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(f(a, x), y) → F(y, f(x, f(a, f(h(a), a))))
F(f(a, x), y) → F(a, f(h(a), a))
F(f(a, x), y) → F(x, f(a, f(h(a), a)))
F(f(a, x), y) → F(h(a), a)

The TRS R consists of the following rules:

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

The set Q consists of the following terms:

f(f(a, x0), x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F(f(a, x), y) → F(y, f(x, f(a, f(h(a), a))))
F(f(a, x), y) → F(a, f(h(a), a))
F(f(a, x), y) → F(x, f(a, f(h(a), a)))
F(f(a, x), y) → F(h(a), a)

The TRS R consists of the following rules:

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

The set Q consists of the following terms:

f(f(a, x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ ForwardInstantiation

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

F(f(a, x), y) → F(y, f(x, f(a, f(h(a), a))))
F(f(a, x), y) → F(x, f(a, f(h(a), a)))

The TRS R consists of the following rules:

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

The set Q consists of the following terms:

f(f(a, x0), x1)

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(f(a, x), y) → F(x, f(a, f(h(a), a))) we obtained the following new rules:

F(f(a, f(a, y_0)), x1) → F(f(a, y_0), f(a, f(h(a), a)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ MNOCProof

                  ↳ SemLabProof

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

F(f(a, x), y) → F(y, f(x, f(a, f(h(a), a))))
F(f(a, f(a, y_0)), x1) → F(f(a, y_0), f(a, f(h(a), a)))

The TRS R consists of the following rules:

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

The set Q consists of the following terms:

f(f(a, x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ MNOCProof

                    ↳ QDP

                  ↳ SemLabProof

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

F(f(a, x), y) → F(y, f(x, f(a, f(h(a), a))))
F(f(a, f(a, y_0)), x1) → F(f(a, y_0), f(a, f(h(a), a)))

The TRS R consists of the following rules:

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

Q is empty.
We have to consider all (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.a: 1
f: 0
h: 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.0-1(f.1-1(a., x), y) → F.1-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), x1) → F.0-0(f.1-1(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-1(f.1-0(a., x), y) → F.1-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-1(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-1(f.1-0(a., f.1-1(a., y_0)), x1) → F.0-0(f.1-1(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-1(a., x), y) → F.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-1(a., x), y) → f.1-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-0(a., x), y) → f.1-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-0(f.1-1(a., x), y) → f.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ MNOCProof

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

F.0-1(f.1-1(a., x), y) → F.1-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), x1) → F.0-0(f.1-1(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-1(f.1-0(a., x), y) → F.1-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-1(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-1(f.1-0(a., f.1-1(a., y_0)), x1) → F.0-0(f.1-1(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-1(a., x), y) → F.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-1(a., x), y) → f.1-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-0(a., x), y) → f.1-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-0(f.1-1(a., x), y) → f.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ MNOCProof

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

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

F.0-0(f.1-0(a., f.1-1(a., y_0)), x1) → F.0-0(f.1-1(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-1(a., x), y) → F.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-1(a., x), y) → f.1-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-0(a., x), y) → f.1-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-0(f.1-1(a., x), y) → f.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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.0-0(f.1-1(a., x), y) → F.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The following rules are removed from R:

f.0-0(f.1-1(a., x), y) → f.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-0(x₁, x₂)) = x₁ + x₂
POL(a.) = 0
POL(f.0-0(x₁, x₂)) = x₁ + x₂
POL(f.0-1(x₁, x₂)) = x₁ + x₂
POL(f.1-0(x₁, x₂)) = x₁ + x₂
POL(f.1-1(x₁, x₂)) = 1 + x₁ + x₂
POL(h.1(x₁)) = x₁

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ MNOCProof

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

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

F.0-0(f.1-0(a., f.1-1(a., y_0)), x1) → F.0-0(f.1-1(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ MNOCProof

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

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

F.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The remaining pairs can at least be oriented weakly.

F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

Used ordering: Polynomial interpretation [25]:

POL(F.0-0(x₁, x₂)) = x₁ + x₂
POL(a.) = 0
POL(f.0-0(x₁, x₂)) = 0
POL(f.0-1(x₁, x₂)) = 0
POL(f.1-0(x₁, x₂)) = 1 + x₂
POL(h.1(x₁)) = 0

The following usable rules [17] were oriented:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ MNOCProof

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ UsableRulesReductionPairsProof

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

F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

POL(F.0-0(x₁, x₂)) = x₁ + x₂
POL(a.) = 0
POL(f.0-1(x₁, x₂)) = x₁ + x₂
POL(f.1-0(x₁, x₂)) = x₁ + x₂
POL(h.1(x₁)) = x₁

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ MNOCProof

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ UsableRulesReductionPairsProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

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

F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

R is empty.
The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(F.0-0(x₁, x₂)) = x₁
POL(a.) = 1
POL(f.0-1(x₁, x₂)) = 0
POL(f.1-0(x₁, x₂)) = 1 + x₁ + x₂
POL(h.1(x₁)) = 0

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ MNOCProof

                  ↳ SemLabProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ UsableRulesReductionPairsProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ QDPOrderProof

                                    ↳ QDP

                                      ↳ UsableRulesReductionPairsProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., 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.