Termination w.r.t. Q proof of /tmp/tmpruUuH2/33.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(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, 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(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, 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:

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

g(e(x), e(y)) → e(g(x, y))

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:

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

The TRS R consists of the following rules:

h(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, 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(z, e(x)) → H(c(z), d(z, x)) we obtained the following new rules:

H(c(y_0), e(x1)) → H(c(c(y_0)), d(c(y_0), x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ Instantiation

              ↳ QDP

                ↳ Instantiation

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

H(c(y_0), e(x1)) → H(c(c(y_0)), d(c(y_0), x1))

The TRS R consists of the following rules:

h(z, e(x)) → h(c(z), d(z, x))
d(z, g(0, 0)) → e(0)
d(z, g(x, y)) → g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) → g(d(c(z), g(x, y)), d(z, g(x, 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(c(y_0), e(x1)) → H(c(c(y_0)), d(c(y_0), x1)) we obtained the following new rules:

H(c(c(y_0)), e(x1)) → H(c(c(c(y_0))), d(c(c(y_0)), x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ Instantiation

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

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

H(c(c(y_0)), e(x1)) → H(c(c(c(y_0))), d(c(c(y_0)), x1))

The TRS R consists of the following rules:

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

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