Termination w.r.t. Q proof of /tmp/tmpzXEpgH/test10.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:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Q is empty.

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(0, Y) → 0
g(X, s(Y)) → g(X, Y)

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

g(0, Y) → 0

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(f(x₁, x₂, x₃)) = x₁ + x₂ + 2·x₃
POL(g(x₁, x₂)) = 2 + x₁ + x₂
POL(h(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

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

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

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:

G(X, s(Y)) → G(X, Y)
H(X, Z) → F(X, s(X), Z)
F(X, Y, g(X, Y)) → H(0, g(X, Y))

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

G(X, s(Y)) → G(X, Y)
H(X, Z) → F(X, s(X), Z)
F(X, Y, g(X, Y)) → H(0, g(X, Y))

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

G(X, s(Y)) → G(X, Y)

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

              ↳ QDP

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

G(X, s(Y)) → G(X, Y)

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:

G(X, s(Y)) → G(X, Y)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

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

H(X, Z) → F(X, s(X), Z)
F(X, Y, g(X, Y)) → H(0, g(X, Y))

The TRS R consists of the following rules:

h(X, Z) → f(X, s(X), Z)
f(X, Y, g(X, Y)) → h(0, g(X, Y))
g(X, s(Y)) → g(X, Y)

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.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(F(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(H(x₁, x₂)) = 2·x₁ + x₂
POL(g(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(s(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ Instantiation

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

H(X, Z) → F(X, s(X), Z)
F(X, Y, g(X, Y)) → H(0, g(X, Y))

The TRS R consists of the following rules:

g(X, s(Y)) → g(X, Y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule H(X, Z) → F(X, s(X), Z) we obtained the following new rules:

H(0, y_2) → F(0, s(0), y_2)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

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

F(X, Y, g(X, Y)) → H(0, g(X, Y))
H(0, y_2) → F(0, s(0), y_2)

The TRS R consists of the following rules:

g(X, s(Y)) → g(X, Y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(X, Y, g(X, Y)) → H(0, g(X, Y)) we obtained the following new rules:

F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

                          ↳ QDP

                            ↳ RuleRemovalProof

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

H(0, y_2) → F(0, s(0), y_2)
F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))

The TRS R consists of the following rules:

g(X, s(Y)) → g(X, Y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

g(X, s(Y)) → g(X, Y)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1
POL(F(x₁, x₂, x₃)) = 2·x₁ + x₂ + x₃
POL(H(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(g(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(s(x₁)) = 1 + x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ Instantiation

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

H(0, y_2) → F(0, s(0), y_2)
F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule H(0, y_2) → F(0, s(0), y_2) we obtained the following new rules:

H(0, g(0, s(0))) → F(0, s(0), g(0, s(0)))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesReductionPairsProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ NonTerminationProof

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

H(0, g(0, s(0))) → F(0, s(0), g(0, s(0)))
F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))

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 narrowing to the left:

The TRS P consists of the following rules:

H(0, g(0, s(0))) → F(0, s(0), g(0, s(0)))
F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))

The TRS R consists of the following rules:none

s = F(0, s(0), g(0, s(0))) evaluates to t =F(0, s(0), g(0, s(0)))

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

F(0, s(0), g(0, s(0))) → H(0, g(0, s(0)))
with rule F(0, s(0), g(0, s(0))) → H(0, g(0, s(0))) at position [] and matcher [ ]

H(0, g(0, s(0))) → F(0, s(0), g(0, s(0)))
with rule H(0, g(0, s(0))) → F(0, s(0), g(0, s(0)))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.