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

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:

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)

The set Q consists of the following terms:

h(x0, x1)
g(0, x0)
f(x0, x1, g(x0, x1))
g(x0, s(x1))

↳ 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)

The set Q consists of the following terms:

h(x0, x1)
g(0, x0)
f(x0, x1, g(x0, x1))
g(x0, s(x1))

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)

The set Q consists of the following terms:

h(x0, x1)
g(0, x0)
f(x0, x1, g(x0, x1))
g(x0, s(x1))

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)

The set Q consists of the following terms:

h(x0, x1)
g(0, x0)
f(x0, x1, g(x0, x1))
g(x0, s(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:

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)

The set Q consists of the following terms:

h(x0, x1)
g(0, x0)
f(x0, x1, g(x0, x1))
g(x0, s(x1))

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)

The set Q consists of the following terms:

h(x0, x1)
g(0, x0)
f(x0, x1, g(x0, x1))
g(x0, s(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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

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)

The set Q consists of the following terms:

h(x0, x1)
g(0, x0)
f(x0, x1, g(x0, x1))
g(x0, s(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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

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

G(X, s(Y)) → G(X, Y)

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

h(x0, x1)
g(0, x0)
f(x0, x1, g(x0, x1))
g(x0, s(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.

h(x0, x1)
g(0, x0)
f(x0, x1, g(x0, x1))
g(x0, s(x1))

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

                    ↳ QDP

                      ↳ QDPSizeChangeProof

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