Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/D33SLASH33.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:

h₂(z, e₁(x)) -> h₂(c₁(z), d₂(z, x))
d₂(z, g₂(0, 0)) -> e₁(0)
d₂(z, g₂(x, y)) -> g₂(e₁(x), d₂(z, y))
d₂(c₁(z), g₂(g₂(x, y), 0)) -> g₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
g₂(e₁(x), e₁(y)) -> e₁(g₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

h₂(z, e₁(x)) -> h₂(c₁(z), d₂(z, x))
d₂(z, g₂(0, 0)) -> e₁(0)
d₂(z, g₂(x, y)) -> g₂(e₁(x), d₂(z, y))
d₂(c₁(z), g₂(g₂(x, y), 0)) -> g₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
g₂(e₁(x), e₁(y)) -> e₁(g₂(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:

H₂(z, e₁(x)) -> H₂(c₁(z), d₂(z, x))
D₂(c₁(z), g₂(g₂(x, y), 0)) -> G₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
D₂(z, g₂(x, y)) -> D₂(z, y)
H₂(z, e₁(x)) -> D₂(z, x)
G₂(e₁(x), e₁(y)) -> G₂(x, y)
D₂(c₁(z), g₂(g₂(x, y), 0)) -> D₂(z, g₂(x, y))
D₂(z, g₂(x, y)) -> G₂(e₁(x), d₂(z, y))
D₂(c₁(z), g₂(g₂(x, y), 0)) -> D₂(c₁(z), g₂(x, y))

The TRS R consists of the following rules:

h₂(z, e₁(x)) -> h₂(c₁(z), d₂(z, x))
d₂(z, g₂(0, 0)) -> e₁(0)
d₂(z, g₂(x, y)) -> g₂(e₁(x), d₂(z, y))
d₂(c₁(z), g₂(g₂(x, y), 0)) -> g₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
g₂(e₁(x), e₁(y)) -> e₁(g₂(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:

H₂(z, e₁(x)) -> H₂(c₁(z), d₂(z, x))
D₂(c₁(z), g₂(g₂(x, y), 0)) -> G₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
D₂(z, g₂(x, y)) -> D₂(z, y)
H₂(z, e₁(x)) -> D₂(z, x)
G₂(e₁(x), e₁(y)) -> G₂(x, y)
D₂(c₁(z), g₂(g₂(x, y), 0)) -> D₂(z, g₂(x, y))
D₂(z, g₂(x, y)) -> G₂(e₁(x), d₂(z, y))
D₂(c₁(z), g₂(g₂(x, y), 0)) -> D₂(c₁(z), g₂(x, y))

The TRS R consists of the following rules:

h₂(z, e₁(x)) -> h₂(c₁(z), d₂(z, x))
d₂(z, g₂(0, 0)) -> e₁(0)
d₂(z, g₂(x, y)) -> g₂(e₁(x), d₂(z, y))
d₂(c₁(z), g₂(g₂(x, y), 0)) -> g₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
g₂(e₁(x), e₁(y)) -> e₁(g₂(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

G₂(e₁(x), e₁(y)) -> G₂(x, y)

The TRS R consists of the following rules:

h₂(z, e₁(x)) -> h₂(c₁(z), d₂(z, x))
d₂(z, g₂(0, 0)) -> e₁(0)
d₂(z, g₂(x, y)) -> g₂(e₁(x), d₂(z, y))
d₂(c₁(z), g₂(g₂(x, y), 0)) -> g₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
g₂(e₁(x), e₁(y)) -> e₁(g₂(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 strictly oriented and are deleted.

G₂(e₁(x), e₁(y)) -> G₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G₂(x1, x2) = G₁(x1)
e₁(x1) = e₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] 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:

h₂(z, e₁(x)) -> h₂(c₁(z), d₂(z, x))
d₂(z, g₂(0, 0)) -> e₁(0)
d₂(z, g₂(x, y)) -> g₂(e₁(x), d₂(z, y))
d₂(c₁(z), g₂(g₂(x, y), 0)) -> g₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
g₂(e₁(x), e₁(y)) -> e₁(g₂(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:

D₂(z, g₂(x, y)) -> D₂(z, y)
D₂(c₁(z), g₂(g₂(x, y), 0)) -> D₂(z, g₂(x, y))
D₂(c₁(z), g₂(g₂(x, y), 0)) -> D₂(c₁(z), g₂(x, y))

The TRS R consists of the following rules:

h₂(z, e₁(x)) -> h₂(c₁(z), d₂(z, x))
d₂(z, g₂(0, 0)) -> e₁(0)
d₂(z, g₂(x, y)) -> g₂(e₁(x), d₂(z, y))
d₂(c₁(z), g₂(g₂(x, y), 0)) -> g₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
g₂(e₁(x), e₁(y)) -> e₁(g₂(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 strictly oriented and are deleted.

D₂(c₁(z), g₂(g₂(x, y), 0)) -> D₂(z, g₂(x, y))

The remaining pairs can at least by weakly be oriented.

D₂(z, g₂(x, y)) -> D₂(z, y)
D₂(c₁(z), g₂(g₂(x, y), 0)) -> D₂(c₁(z), g₂(x, y))

Used ordering: Combined order from the following AFS and order.
D₂(x1, x2) = x1
g₂(x1, x2) = g
c₁(x1) = c₁(x1)
0 = 0
e₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

0 > c₁ > g

The following usable rules [14] were oriented:

g₂(e₁(x), e₁(y)) -> e₁(g₂(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

D₂(z, g₂(x, y)) -> D₂(z, y)
D₂(c₁(z), g₂(g₂(x, y), 0)) -> D₂(c₁(z), g₂(x, y))

The TRS R consists of the following rules:

h₂(z, e₁(x)) -> h₂(c₁(z), d₂(z, x))
d₂(z, g₂(0, 0)) -> e₁(0)
d₂(z, g₂(x, y)) -> g₂(e₁(x), d₂(z, y))
d₂(c₁(z), g₂(g₂(x, y), 0)) -> g₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
g₂(e₁(x), e₁(y)) -> e₁(g₂(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 strictly oriented and are deleted.

D₂(z, g₂(x, y)) -> D₂(z, y)
D₂(c₁(z), g₂(g₂(x, y), 0)) -> D₂(c₁(z), g₂(x, y))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
D₂(x1, x2) = D₂(x1, x2)
g₂(x1, x2) = g₂(x1, x2)
c₁(x1) = x1
0 = 0
e₁(x1) = e

Lexicographic Path Order [19].
Precedence:

e > [D₂, g_2]

The following usable rules [14] were oriented:

g₂(e₁(x), e₁(y)) -> e₁(g₂(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

h₂(z, e₁(x)) -> h₂(c₁(z), d₂(z, x))
d₂(z, g₂(0, 0)) -> e₁(0)
d₂(z, g₂(x, y)) -> g₂(e₁(x), d₂(z, y))
d₂(c₁(z), g₂(g₂(x, y), 0)) -> g₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
g₂(e₁(x), e₁(y)) -> e₁(g₂(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:

H₂(z, e₁(x)) -> H₂(c₁(z), d₂(z, x))

The TRS R consists of the following rules:

h₂(z, e₁(x)) -> h₂(c₁(z), d₂(z, x))
d₂(z, g₂(0, 0)) -> e₁(0)
d₂(z, g₂(x, y)) -> g₂(e₁(x), d₂(z, y))
d₂(c₁(z), g₂(g₂(x, y), 0)) -> g₂(d₂(c₁(z), g₂(x, y)), d₂(z, g₂(x, y)))
g₂(e₁(x), e₁(y)) -> e₁(g₂(x, y))

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