Termination w.r.t. Q proof of /tmp/tmp75GLtx/hydra.trs

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:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(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:

H(e(x), y) → H(d(x, y), s(y))
D(g(g(0, x), y), s(z)) → G(e(x), d(g(g(0, x), y), z))
H(e(x), y) → D(x, y)
D(g(x, y), z) → D(x, z)
D(g(x, y), z) → G(d(x, z), e(y))
G(e(x), e(y)) → G(x, y)
D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

Q is empty.
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:

H(e(x), y) → H(d(x, y), s(y))
D(g(g(0, x), y), s(z)) → G(e(x), d(g(g(0, x), y), z))
H(e(x), y) → D(x, y)
D(g(x, y), z) → D(x, z)
D(g(x, y), z) → G(d(x, z), e(y))
G(e(x), e(y)) → G(x, y)
D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(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 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

G(e(x), e(y)) → G(x, y)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

G(e(x), e(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(e(x), e(y)) → G(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

D(g(x, y), z) → D(x, z)
D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

D(g(x, y), z) → D(x, z)
D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)

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

D(g(x, y), z) → D(x, z)
The graph contains the following edges 1 > 1, 2 >= 2
D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ Instantiation

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

H(e(x), y) → H(d(x, y), s(y))

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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

H(e(x0), s(y_3)) → H(d(x0, s(y_3)), s(s(y_3)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ Instantiation

              ↳ QDP

                ↳ Instantiation

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

H(e(x0), s(y_3)) → H(d(x0, s(y_3)), s(s(y_3)))

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule H(e(x0), s(y_3)) → H(d(x0, s(y_3)), s(s(y_3))) we obtained the following new rules:

H(e(x0), s(s(y_3))) → H(d(x0, s(s(y_3))), s(s(s(y_3))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ Instantiation

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

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

H(e(x0), s(s(y_3))) → H(d(x0, s(s(y_3))), s(s(s(y_3))))

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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