Termination w.r.t. Q proof of TRS/Zantema06hydra.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(e₁(x), y) -> h₂(d₂(x, y), s₁(y))
d₂(g₂(g₂(0, x), y), s₁(z)) -> g₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
d₂(g₂(g₂(0, x), y), 0) -> e₁(y)
d₂(g₂(0, x), y) -> e₁(x)
d₂(g₂(x, y), z) -> g₂(d₂(x, z), e₁(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(e₁(x), y) -> h₂(d₂(x, y), s₁(y))
d₂(g₂(g₂(0, x), y), s₁(z)) -> g₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
d₂(g₂(g₂(0, x), y), 0) -> e₁(y)
d₂(g₂(0, x), y) -> e₁(x)
d₂(g₂(x, y), z) -> g₂(d₂(x, z), e₁(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:

D₂(g₂(g₂(0, x), y), s₁(z)) -> D₂(g₂(g₂(0, x), y), z)
H₂(e₁(x), y) -> D₂(x, y)
D₂(g₂(x, y), z) -> D₂(x, z)
D₂(g₂(g₂(0, x), y), s₁(z)) -> G₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
H₂(e₁(x), y) -> H₂(d₂(x, y), s₁(y))
G₂(e₁(x), e₁(y)) -> G₂(x, y)
D₂(g₂(x, y), z) -> G₂(d₂(x, z), e₁(y))

The TRS R consists of the following rules:

h₂(e₁(x), y) -> h₂(d₂(x, y), s₁(y))
d₂(g₂(g₂(0, x), y), s₁(z)) -> g₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
d₂(g₂(g₂(0, x), y), 0) -> e₁(y)
d₂(g₂(0, x), y) -> e₁(x)
d₂(g₂(x, y), z) -> g₂(d₂(x, z), e₁(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:

D₂(g₂(g₂(0, x), y), s₁(z)) -> D₂(g₂(g₂(0, x), y), z)
H₂(e₁(x), y) -> D₂(x, y)
D₂(g₂(x, y), z) -> D₂(x, z)
D₂(g₂(g₂(0, x), y), s₁(z)) -> G₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
H₂(e₁(x), y) -> H₂(d₂(x, y), s₁(y))
G₂(e₁(x), e₁(y)) -> G₂(x, y)
D₂(g₂(x, y), z) -> G₂(d₂(x, z), e₁(y))

The TRS R consists of the following rules:

h₂(e₁(x), y) -> h₂(d₂(x, y), s₁(y))
d₂(g₂(g₂(0, x), y), s₁(z)) -> g₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
d₂(g₂(g₂(0, x), y), 0) -> e₁(y)
d₂(g₂(0, x), y) -> e₁(x)
d₂(g₂(x, y), z) -> g₂(d₂(x, z), e₁(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(e₁(x), y) -> h₂(d₂(x, y), s₁(y))
d₂(g₂(g₂(0, x), y), s₁(z)) -> g₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
d₂(g₂(g₂(0, x), y), 0) -> e₁(y)
d₂(g₂(0, x), y) -> e₁(x)
d₂(g₂(x, y), z) -> g₂(d₂(x, z), e₁(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 oriented strictly and are deleted.

G₂(e₁(x), e₁(y)) -> G₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(G₂(x₁, x₂)) = 2·x₁ + x₂
POL(e₁(x₁)) = 2 + 2·x₁

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(e₁(x), y) -> h₂(d₂(x, y), s₁(y))
d₂(g₂(g₂(0, x), y), s₁(z)) -> g₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
d₂(g₂(g₂(0, x), y), 0) -> e₁(y)
d₂(g₂(0, x), y) -> e₁(x)
d₂(g₂(x, y), z) -> g₂(d₂(x, z), e₁(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(g₂(g₂(0, x), y), s₁(z)) -> D₂(g₂(g₂(0, x), y), z)
D₂(g₂(x, y), z) -> D₂(x, z)

The TRS R consists of the following rules:

h₂(e₁(x), y) -> h₂(d₂(x, y), s₁(y))
d₂(g₂(g₂(0, x), y), s₁(z)) -> g₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
d₂(g₂(g₂(0, x), y), 0) -> e₁(y)
d₂(g₂(0, x), y) -> e₁(x)
d₂(g₂(x, y), z) -> g₂(d₂(x, z), e₁(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 oriented strictly and are deleted.

D₂(g₂(g₂(0, x), y), s₁(z)) -> D₂(g₂(g₂(0, x), y), z)
D₂(g₂(x, y), z) -> D₂(x, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(D₂(x₁, x₂)) = x₁ + 2·x₂
POL(g₂(x₁, x₂)) = 1 + 2·x₁
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ 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:

h₂(e₁(x), y) -> h₂(d₂(x, y), s₁(y))
d₂(g₂(g₂(0, x), y), s₁(z)) -> g₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
d₂(g₂(g₂(0, x), y), 0) -> e₁(y)
d₂(g₂(0, x), y) -> e₁(x)
d₂(g₂(x, y), z) -> g₂(d₂(x, z), e₁(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(e₁(x), y) -> H₂(d₂(x, y), s₁(y))

The TRS R consists of the following rules:

h₂(e₁(x), y) -> h₂(d₂(x, y), s₁(y))
d₂(g₂(g₂(0, x), y), s₁(z)) -> g₂(e₁(x), d₂(g₂(g₂(0, x), y), z))
d₂(g₂(g₂(0, x), y), 0) -> e₁(y)
d₂(g₂(0, x), y) -> e₁(x)
d₂(g₂(x, y), z) -> g₂(d₂(x, z), e₁(y))
g₂(e₁(x), e₁(y)) -> e₁(g₂(x, y))

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