Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/SK90SLASH2.03.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

Q is empty.

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

MINUS₁(h₁(x)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(y)

The TRS R consists of the following rules:

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

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

MINUS₁(h₁(x)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(y)

The TRS R consists of the following rules:

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

Q is empty.
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.

MINUS₁(f₂(x, y)) -> MINUS₁(x)
MINUS₁(f₂(x, y)) -> MINUS₁(y)

Used argument filtering: MINUS₁(x1) = x1
h₁(x1) = x1
f₂(x1, x2) = f₂(x1, x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

MINUS₁(h₁(x)) -> MINUS₁(x)

The TRS R consists of the following rules:

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

Q is empty.
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.

MINUS₁(h₁(x)) -> MINUS₁(x)

Used argument filtering: MINUS₁(x1) = x1
h₁(x1) = h₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ PisEmptyProof

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

minus₁(minus₁(x)) -> x
minus₁(h₁(x)) -> h₁(minus₁(x))
minus₁(f₂(x, y)) -> f₂(minus₁(y), minus₁(x))

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.