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

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f₃(t, x, y) -> f₃(g₂(x, y), x, s₁(y))
g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)
f₃(t, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

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

f₃(t, x, y) -> f₃(g₂(x, y), x, s₁(y))
g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)
f₃(t, 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:

G₂(s₁(x), s₁(y)) -> G₂(x, y)
F₃(t, x, y) -> F₃(g₂(x, y), x, s₁(y))
F₃(t, x, y) -> G₂(x, y)

The TRS R consists of the following rules:

f₃(t, x, y) -> f₃(g₂(x, y), x, s₁(y))
g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)
f₃(t, x0, x1)

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:

G₂(s₁(x), s₁(y)) -> G₂(x, y)
F₃(t, x, y) -> F₃(g₂(x, y), x, s₁(y))
F₃(t, x, y) -> G₂(x, y)

The TRS R consists of the following rules:

f₃(t, x, y) -> f₃(g₂(x, y), x, s₁(y))
g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)
f₃(t, 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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

G₂(s₁(x), s₁(y)) -> G₂(x, y)

The TRS R consists of the following rules:

f₃(t, x, y) -> f₃(g₂(x, y), x, s₁(y))
g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)
f₃(t, 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 [13] we can delete all non-usable rules [14] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

G₂(s₁(x), s₁(y)) -> G₂(x, y)

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

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)
f₃(t, 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.

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)
f₃(t, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

G₂(s₁(x), s₁(y)) -> G₂(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [16] together with the size-change analysis [27] 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:

G₂(s₁(x), s₁(y)) -> G₂(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

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

F₃(t, x, y) -> F₃(g₂(x, y), x, s₁(y))

The TRS R consists of the following rules:

f₃(t, x, y) -> f₃(g₂(x, y), x, s₁(y))
g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)
f₃(t, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

            ↳ UsableRulesProof

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

F₃(t, x, y) -> F₃(g₂(x, y), x, s₁(y))

The TRS R consists of the following rules:

g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)
f₃(t, 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.

f₃(t, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

            ↳ UsableRulesProof

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

F₃(t, x, y) -> F₃(g₂(x, y), x, s₁(y))

The TRS R consists of the following rules:

g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule F₃(t, x, y) -> F₃(g₂(x, y), x, s₁(y)) at position [0] we obtained the following new rules:

F₃(t, s₁(x0), 0) -> F₃(t, s₁(x0), s₁(0))
F₃(t, s₁(x0), s₁(x1)) -> F₃(g₂(x0, x1), s₁(x0), s₁(s₁(x1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

            ↳ UsableRulesProof

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

F₃(t, s₁(x0), 0) -> F₃(t, s₁(x0), s₁(0))
F₃(t, s₁(x0), s₁(x1)) -> F₃(g₂(x0, x1), s₁(x0), s₁(s₁(x1)))

The TRS R consists of the following rules:

g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

            ↳ UsableRulesProof

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

F₃(t, s₁(x0), s₁(x1)) -> F₃(g₂(x0, x1), s₁(x0), s₁(s₁(x1)))

The TRS R consists of the following rules:

g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule F₃(t, s₁(x0), s₁(x1)) -> F₃(g₂(x0, x1), s₁(x0), s₁(s₁(x1))) we obtained the following new rules:

F₃(t, s₁(z0), s₁(s₁(z1))) -> F₃(g₂(z0, s₁(z1)), s₁(z0), s₁(s₁(s₁(z1))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

            ↳ UsableRulesProof

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

F₃(t, s₁(z0), s₁(s₁(z1))) -> F₃(g₂(z0, s₁(z1)), s₁(z0), s₁(s₁(s₁(z1))))

The TRS R consists of the following rules:

g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

F₃(t, x, y) -> F₃(g₂(x, y), x, s₁(y))

The TRS R consists of the following rules:

g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)
f₃(t, 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.

f₃(t, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

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

F₃(t, x, y) -> F₃(g₂(x, y), x, s₁(y))

The TRS R consists of the following rules:

g₂(s₁(x), 0) -> t
g₂(s₁(x), s₁(y)) -> g₂(x, y)

The set Q consists of the following terms:

g₂(s₁(x0), s₁(x1))
g₂(s₁(x0), 0)

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