Termination w.r.t. Q proof of TRS/D3333.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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

H(z, e(x)) → H(c(z), d(z, x))
D(c(z), g(g(x, y), 0)) → G(d(c(z), g(x, y)), d(z, g(x, y)))
D(z, g(x, y)) → D(z, y)
H(z, e(x)) → D(z, x)
G(e(x), e(y)) → G(x, y)
D(c(z), g(g(x, y), 0)) → D(z, g(x, y))
D(z, g(x, y)) → G(e(x), d(z, y))
D(c(z), g(g(x, y), 0)) → D(c(z), 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 deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

H(z, e(x)) → H(c(z), d(z, x))
D(c(z), g(g(x, y), 0)) → G(d(c(z), g(x, y)), d(z, g(x, y)))
D(z, g(x, y)) → D(z, y)
H(z, e(x)) → D(z, x)
G(e(x), e(y)) → G(x, y)
D(c(z), g(g(x, y), 0)) → D(z, g(x, y))
D(z, g(x, y)) → G(e(x), d(z, y))
D(c(z), g(g(x, y), 0)) → D(c(z), 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.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ 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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

G(e(x), e(y)) → G(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1, x2) = G(x2)
e(x1) = e(x1)

Lexicographic Path Order [19].
Precedence:

e₁ > G₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

D(z, g(x, y)) → D(z, y)
D(c(z), g(g(x, y), 0)) → D(z, g(x, y))
D(c(z), g(g(x, y), 0)) → D(c(z), 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 use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
D(x1, x2) = D(x1, x2)
g(x1, x2) = g(x1, x2)
c(x1) = x1
0 = 0
e(x1) = e(x1)

Lexicographic Path Order [19].
Precedence:

0 > g₂ > e₁ > D₂

The following usable rules [14] were oriented:

g(e(x), e(y)) → e(g(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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.