Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/AG01SLASHHASH3.36.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:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(x0))

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

G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
F₁(s₁(x)) -> G₁(f₁(x))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
G₁(s₁(x)) -> MINUS₂(s₁(x), f₁(g₁(x)))
F₁(s₁(x)) -> MINUS₂(s₁(x), g₁(f₁(x)))
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(x0))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
F₁(s₁(x)) -> G₁(f₁(x))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
G₁(s₁(x)) -> MINUS₂(s₁(x), f₁(g₁(x)))
F₁(s₁(x)) -> MINUS₂(s₁(x), g₁(f₁(x)))
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

Used argument filtering: MINUS₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

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

G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
F₁(s₁(x)) -> G₁(f₁(x))
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> minus₂(s₁(x), g₁(f₁(x)))
g₁(0) -> 0
g₁(s₁(x)) -> minus₂(s₁(x), f₁(g₁(x)))

The set Q consists of the following terms:

minus₂(x0, 0)
minus₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(x0))

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