Termination w.r.t. Q proof of /tmp/tmpUT3qj0/z28.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(0, x), 1) → f(g(f(x, x)), x)
f(g(x), y) → g(f(x, y))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

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

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

The set Q consists of the following terms:

f(f(0, x0), 1)
f(g(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(0, x), 1) → F(g(f(x, x)), x)
F(f(0, x), 1) → F(x, x)
F(g(x), y) → F(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(0, x0), 1)
f(g(x0), x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ ForwardInstantiation

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(0, x0), 1)
f(g(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(f(0, x), 1) → F(x, x) we obtained the following new rules:

F(f(0, g(y_0)), 1) → F(g(y_0), g(y_0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ ForwardInstantiation

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

F(f(0, x), 1) → F(g(f(x, x)), x)
F(g(x), y) → F(x, y)
F(f(0, g(y_0)), 1) → F(g(y_0), g(y_0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(0, x0), 1)
f(g(x0), x1)

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

F(g(f(0, g(y_0))), 1) → F(f(0, g(y_0)), 1)
F(g(g(y_0)), x1) → F(g(y_0), x1)
F(g(f(0, y_0)), 1) → F(f(0, y_0), 1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ Narrowing

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

F(f(0, x), 1) → F(g(f(x, x)), x)
F(f(0, g(y_0)), 1) → F(g(y_0), g(y_0))
F(g(f(0, g(y_0))), 1) → F(f(0, g(y_0)), 1)
F(g(f(0, y_0)), 1) → F(f(0, y_0), 1)
F(g(g(y_0)), x1) → F(g(y_0), x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(0, x0), 1)
f(g(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(f(0, x), 1) → F(g(f(x, x)), x) at position [] we obtained the following new rules:

F(f(0, g(x0)), 1) → F(g(g(f(x0, g(x0)))), g(x0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ ForwardInstantiation

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

F(f(0, g(x0)), 1) → F(g(g(f(x0, g(x0)))), g(x0))
F(f(0, g(y_0)), 1) → F(g(y_0), g(y_0))
F(g(f(0, g(y_0))), 1) → F(f(0, g(y_0)), 1)
F(g(g(y_0)), x1) → F(g(y_0), x1)
F(g(f(0, y_0)), 1) → F(f(0, y_0), 1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(0, x0), 1)
f(g(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(g(f(0, y_0)), 1) → F(f(0, y_0), 1) we obtained the following new rules:

F(g(f(0, g(y_0))), 1) → F(f(0, g(y_0)), 1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ ForwardInstantiation

                        ↳ QDP

                          ↳ QDPOrderProof

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

F(f(0, g(x0)), 1) → F(g(g(f(x0, g(x0)))), g(x0))
F(f(0, g(y_0)), 1) → F(g(y_0), g(y_0))
F(g(f(0, g(y_0))), 1) → F(f(0, g(y_0)), 1)
F(g(g(y_0)), x1) → F(g(y_0), x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(0, x0), 1)
f(g(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(f(0, g(x0)), 1) → F(g(g(f(x0, g(x0)))), g(x0))
F(f(0, g(y_0)), 1) → F(g(y_0), g(y_0))

The remaining pairs can at least be oriented weakly.

F(g(f(0, g(y_0))), 1) → F(f(0, g(y_0)), 1)
F(g(g(y_0)), x1) → F(g(y_0), x1)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(1) = 1
POL(F(x₁, x₂)) = x₂
POL(f(x₁, x₂)) = 0
POL(g(x₁)) = 0

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ ForwardInstantiation

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

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

F(g(f(0, g(y_0))), 1) → F(f(0, g(y_0)), 1)
F(g(g(y_0)), x1) → F(g(y_0), x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(0, x0), 1)
f(g(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

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ ForwardInstantiation

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ UsableRulesProof

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

F(g(g(y_0)), x1) → F(g(y_0), x1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(0, x0), 1)
f(g(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ ForwardInstantiation

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

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

F(g(g(y_0)), x1) → F(g(y_0), x1)

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

f(f(0, x0), 1)
f(g(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(f(0, x0), 1)
f(g(x0), x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ ForwardInstantiation

            ↳ QDP

              ↳ ForwardInstantiation

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ ForwardInstantiation

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

                              ↳ DependencyGraphProof

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ QReductionProof

                                        ↳ QDP

                                          ↳ QDPSizeChangeProof

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

F(g(g(y_0)), x1) → F(g(y_0), x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

F(g(g(y_0)), x1) → F(g(y_0), x1)
The graph contains the following edges 1 > 1, 2 >= 2