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

            ↳ 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 [15].

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: Polynomial interpretation [25,35]:

POL(e(x₁)) = 5/4 + (15/4)x_1
POL(G(x₁, x₂)) = x_1 + (13/4)x_2

The value of delta used in the strict ordering is 85/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(c(x₁)) = 1/4 + (11/4)x_1
POL(g(x₁, x₂)) = 1/2 + x_1 + (4)x_2
POL(e(x₁)) = 2 + (5/4)x_1
POL(D(x₁, x₂)) = (1/4)x_1 + (3)x_2
POL(0) = 4

The value of delta used in the strict ordering is 3/2.
The following usable rules [17] were oriented:

g(e(x), e(y)) → e(g(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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

      ↳ 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.