Termination w.r.t. Q proof of /tmp/tmpqZJkx0/4.06.trs

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

The set Q consists of the following terms:

*(x0, +(x1, f(x2)))
*(*(x0, x1), x2)
*(+(x0, x1), x2)

↳ QTRS

  ↳ DependencyPairsProof

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

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

The set Q consists of the following terms:

*(x0, +(x1, f(x2)))
*(*(x0, x1), x2)
*(+(x0, x1), x2)

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

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

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

The set Q consists of the following terms:

*(x0, +(x1, f(x2)))
*(*(x0, x1), x2)
*(+(x0, x1), x2)

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:

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

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

The set Q consists of the following terms:

*(x0, +(x1, f(x2)))
*(*(x0, x1), x2)
*(+(x0, x1), x2)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

The set Q consists of the following terms:

*(x0, +(x1, f(x2)))
*(*(x0, x1), x2)
*(+(x0, x1), x2)

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

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

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

*(x0, +(x1, f(x2)))
*(*(x0, x1), x2)
*(+(x0, x1), x2)

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.

*(x0, +(x1, f(x2)))
*(*(x0, x1), x2)
*(+(x0, x1), x2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule *¹(x, +(y, f(z))) → *¹(g(x, z), +(y, y)) we obtained the following new rules:

*¹(g(z0, z2), +(f(x2), f(x2))) → *¹(g(g(z0, z2), x2), +(f(x2), f(x2)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

          ↳ QDP

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

*¹(g(z0, z2), +(f(x2), f(x2))) → *¹(g(g(z0, z2), x2), +(f(x2), f(x2)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule *¹(g(z0, z2), +(f(x2), f(x2))) → *¹(g(g(z0, z2), x2), +(f(x2), f(x2))) we obtained the following new rules:

*¹(g(g(z0, z1), z2), +(f(z2), f(z2))) → *¹(g(g(g(z0, z1), z2), z2), +(f(z2), f(z2)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

                          ↳ QDP

                            ↳ NonTerminationProof

          ↳ QDP

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

*¹(g(g(z0, z1), z2), +(f(z2), f(z2))) → *¹(g(g(g(z0, z1), z2), z2), +(f(z2), f(z2)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

*¹(g(g(z0, z1), z2), +(f(z2), f(z2))) → *¹(g(g(g(z0, z1), z2), z2), +(f(z2), f(z2)))

The TRS R consists of the following rules:none

s = *¹(g(g(z0, z1), z2), +(f(z2), f(z2))) evaluates to t =*¹(g(g(g(z0, z1), z2), z2), +(f(z2), f(z2)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [z1 / z2, z0 / g(z0, z1)]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from *^1(g(g(z0, z1), z2), +(f(z2), f(z2))) to *^1(g(g(g(z0, z1), z2), z2), +(f(z2), f(z2))).

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

*¹(+(x, y), z) → *¹(x, z)
*¹(*(x, y), z) → *¹(y, z)
*¹(+(x, y), z) → *¹(y, z)

The TRS R consists of the following rules:

*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, f(z))) → *(g(x, z), +(y, y))

The set Q consists of the following terms:

*(x0, +(x1, f(x2)))
*(*(x0, x1), x2)
*(+(x0, x1), x2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

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

*¹(+(x, y), z) → *¹(x, z)
*¹(*(x, y), z) → *¹(y, z)
*¹(+(x, y), z) → *¹(y, z)

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

*(x0, +(x1, f(x2)))
*(*(x0, x1), x2)
*(+(x0, x1), x2)

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:

*¹(+(x, y), z) → *¹(x, z)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(*(x, y), z) → *¹(y, z)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(+(x, y), z) → *¹(y, z)
The graph contains the following edges 1 > 1, 2 >= 2