Termination w.r.t. Q proof of TRS/SK902.36.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:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(x, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(x, z))

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:

IMPLIES(x, or(y, z)) → IMPLIES(x, z)
IMPLIES(not(x), or(y, z)) → IMPLIES(y, or(x, z))

The TRS R consists of the following rules:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(x, z))

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:

IMPLIES(x, or(y, z)) → IMPLIES(x, z)
IMPLIES(not(x), or(y, z)) → IMPLIES(y, or(x, z))

The TRS R consists of the following rules:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(x, z))

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.

IMPLIES(x, or(y, z)) → IMPLIES(x, z)

The remaining pairs can at least be oriented weakly.

IMPLIES(not(x), or(y, z)) → IMPLIES(y, or(x, z))

Used ordering: Combined order from the following AFS and order.
IMPLIES(x1, x2) = IMPLIES(x2)
or(x1, x2) = or(x2)
not(x1) = x1

Recursive path order with status [2].
Quasi-Precedence:

IMPLIES₁ > or₁

Status:

or₁: multiset
IMPLIES₁: multiset

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:

IMPLIES(not(x), or(y, z)) → IMPLIES(y, or(x, z))

The TRS R consists of the following rules:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(x, z))

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.

IMPLIES(not(x), or(y, z)) → IMPLIES(y, or(x, z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
IMPLIES(x1, x2) = IMPLIES(x1, x2)
not(x1) = not(x1)
or(x1, x2) = or(x1)

Recursive path order with status [2].
Quasi-Precedence:

[not₁, or_1] > IMPLIES₂

Status:

IMPLIES₂: multiset
or₁: multiset
not₁: multiset

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:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(x, z))

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.