Termination w.r.t. Q proof of TRS/Thiemannidentity.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:

g₂(x, 0) -> 0
g₂(d, s₁(x)) -> s₁(s₁(g₂(d, x)))
g₂(h, s₁(0)) -> 0
g₂(h, s₁(s₁(x))) -> s₁(g₂(h, x))
double₁(x) -> g₂(d, x)
half₁(x) -> g₂(h, x)
f₂(s₁(x), y) -> f₂(half₁(s₁(x)), double₁(y))
f₂(s₁(0), y) -> y
id₁(x) -> f₂(x, s₁(0))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g₂(x, 0) -> 0
g₂(d, s₁(x)) -> s₁(s₁(g₂(d, x)))
g₂(h, s₁(0)) -> 0
g₂(h, s₁(s₁(x))) -> s₁(g₂(h, x))
double₁(x) -> g₂(d, x)
half₁(x) -> g₂(h, x)
f₂(s₁(x), y) -> f₂(half₁(s₁(x)), double₁(y))
f₂(s₁(0), y) -> y
id₁(x) -> f₂(x, s₁(0))

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:

G₂(h, s₁(s₁(x))) -> G₂(h, x)
G₂(d, s₁(x)) -> G₂(d, x)
ID₁(x) -> F₂(x, s₁(0))
F₂(s₁(x), y) -> DOUBLE₁(y)
F₂(s₁(x), y) -> F₂(half₁(s₁(x)), double₁(y))
HALF₁(x) -> G₂(h, x)
F₂(s₁(x), y) -> HALF₁(s₁(x))
DOUBLE₁(x) -> G₂(d, x)

The TRS R consists of the following rules:

g₂(x, 0) -> 0
g₂(d, s₁(x)) -> s₁(s₁(g₂(d, x)))
g₂(h, s₁(0)) -> 0
g₂(h, s₁(s₁(x))) -> s₁(g₂(h, x))
double₁(x) -> g₂(d, x)
half₁(x) -> g₂(h, x)
f₂(s₁(x), y) -> f₂(half₁(s₁(x)), double₁(y))
f₂(s₁(0), y) -> y
id₁(x) -> f₂(x, s₁(0))

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:

G₂(h, s₁(s₁(x))) -> G₂(h, x)
G₂(d, s₁(x)) -> G₂(d, x)
ID₁(x) -> F₂(x, s₁(0))
F₂(s₁(x), y) -> DOUBLE₁(y)
F₂(s₁(x), y) -> F₂(half₁(s₁(x)), double₁(y))
HALF₁(x) -> G₂(h, x)
F₂(s₁(x), y) -> HALF₁(s₁(x))
DOUBLE₁(x) -> G₂(d, x)

The TRS R consists of the following rules:

g₂(x, 0) -> 0
g₂(d, s₁(x)) -> s₁(s₁(g₂(d, x)))
g₂(h, s₁(0)) -> 0
g₂(h, s₁(s₁(x))) -> s₁(g₂(h, x))
double₁(x) -> g₂(d, x)
half₁(x) -> g₂(h, x)
f₂(s₁(x), y) -> f₂(half₁(s₁(x)), double₁(y))
f₂(s₁(0), y) -> y
id₁(x) -> f₂(x, s₁(0))

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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

G₂(h, s₁(s₁(x))) -> G₂(h, x)

The TRS R consists of the following rules:

g₂(x, 0) -> 0
g₂(d, s₁(x)) -> s₁(s₁(g₂(d, x)))
g₂(h, s₁(0)) -> 0
g₂(h, s₁(s₁(x))) -> s₁(g₂(h, x))
double₁(x) -> g₂(d, x)
half₁(x) -> g₂(h, x)
f₂(s₁(x), y) -> f₂(half₁(s₁(x)), double₁(y))
f₂(s₁(0), y) -> y
id₁(x) -> f₂(x, s₁(0))

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(h, s₁(s₁(x))) -> G₂(h, x)

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

POL(G₂(x₁, x₂)) = 2·x₂
POL(h) = 0
POL(s₁(x₁)) = 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:

g₂(x, 0) -> 0
g₂(d, s₁(x)) -> s₁(s₁(g₂(d, x)))
g₂(h, s₁(0)) -> 0
g₂(h, s₁(s₁(x))) -> s₁(g₂(h, x))
double₁(x) -> g₂(d, x)
half₁(x) -> g₂(h, x)
f₂(s₁(x), y) -> f₂(half₁(s₁(x)), double₁(y))
f₂(s₁(0), y) -> y
id₁(x) -> f₂(x, s₁(0))

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:

G₂(d, s₁(x)) -> G₂(d, x)

The TRS R consists of the following rules:

g₂(x, 0) -> 0
g₂(d, s₁(x)) -> s₁(s₁(g₂(d, x)))
g₂(h, s₁(0)) -> 0
g₂(h, s₁(s₁(x))) -> s₁(g₂(h, x))
double₁(x) -> g₂(d, x)
half₁(x) -> g₂(h, x)
f₂(s₁(x), y) -> f₂(half₁(s₁(x)), double₁(y))
f₂(s₁(0), y) -> y
id₁(x) -> f₂(x, s₁(0))

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(d, s₁(x)) -> G₂(d, x)

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

POL(G₂(x₁, x₂)) = 2·x₂
POL(d) = 0
POL(s₁(x₁)) = 1 + 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:

g₂(x, 0) -> 0
g₂(d, s₁(x)) -> s₁(s₁(g₂(d, x)))
g₂(h, s₁(0)) -> 0
g₂(h, s₁(s₁(x))) -> s₁(g₂(h, x))
double₁(x) -> g₂(d, x)
half₁(x) -> g₂(h, x)
f₂(s₁(x), y) -> f₂(half₁(s₁(x)), double₁(y))
f₂(s₁(0), y) -> y
id₁(x) -> f₂(x, s₁(0))

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:

F₂(s₁(x), y) -> F₂(half₁(s₁(x)), double₁(y))

The TRS R consists of the following rules:

g₂(x, 0) -> 0
g₂(d, s₁(x)) -> s₁(s₁(g₂(d, x)))
g₂(h, s₁(0)) -> 0
g₂(h, s₁(s₁(x))) -> s₁(g₂(h, x))
double₁(x) -> g₂(d, x)
half₁(x) -> g₂(h, x)
f₂(s₁(x), y) -> f₂(half₁(s₁(x)), double₁(y))
f₂(s₁(0), y) -> y
id₁(x) -> f₂(x, s₁(0))

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