Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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

      ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

MINUS₁(h₁(x)) -> MINUS₁(x)

Used ordering: Combined order from the following AFS and order.
MINUS₁(x1) = MINUS₁(x1)
h₁(x1) = x1
f₂(x1, x2) = f₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

[MINUS₁, f_2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

MINUS₁(h₁(x)) -> MINUS₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS₁(x1) = MINUS₁(x1)
h₁(x1) = h₁(x1)

Lexicographic Path Order [19].
Precedence:

h₁ > MINUS₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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.