Termination w.r.t. Q proof of TRS/SK902.35.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:

and₂(x, false) -> false
and₂(x, not₁(false)) -> x
not₁(not₁(x)) -> x
implies₂(false, y) -> not₁(false)
implies₂(x, false) -> not₁(x)
implies₂(not₁(x), not₁(y)) -> implies₂(y, and₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

and₂(x, false) -> false
and₂(x, not₁(false)) -> x
not₁(not₁(x)) -> x
implies₂(false, y) -> not₁(false)
implies₂(x, false) -> not₁(x)
implies₂(not₁(x), not₁(y)) -> implies₂(y, and₂(x, y))

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:

IMPLIES₂(not₁(x), not₁(y)) -> IMPLIES₂(y, and₂(x, y))
IMPLIES₂(false, y) -> NOT₁(false)
IMPLIES₂(not₁(x), not₁(y)) -> AND₂(x, y)
IMPLIES₂(x, false) -> NOT₁(x)

The TRS R consists of the following rules:

and₂(x, false) -> false
and₂(x, not₁(false)) -> x
not₁(not₁(x)) -> x
implies₂(false, y) -> not₁(false)
implies₂(x, false) -> not₁(x)
implies₂(not₁(x), not₁(y)) -> implies₂(y, and₂(x, y))

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:

IMPLIES₂(not₁(x), not₁(y)) -> IMPLIES₂(y, and₂(x, y))
IMPLIES₂(false, y) -> NOT₁(false)
IMPLIES₂(not₁(x), not₁(y)) -> AND₂(x, y)
IMPLIES₂(x, false) -> NOT₁(x)

The TRS R consists of the following rules:

and₂(x, false) -> false
and₂(x, not₁(false)) -> x
not₁(not₁(x)) -> x
implies₂(false, y) -> not₁(false)
implies₂(x, false) -> not₁(x)
implies₂(not₁(x), not₁(y)) -> implies₂(y, and₂(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

IMPLIES₂(not₁(x), not₁(y)) -> IMPLIES₂(y, and₂(x, y))

The TRS R consists of the following rules:

and₂(x, false) -> false
and₂(x, not₁(false)) -> x
not₁(not₁(x)) -> x
implies₂(false, y) -> not₁(false)
implies₂(x, false) -> not₁(x)
implies₂(not₁(x), not₁(y)) -> implies₂(y, and₂(x, y))

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.

IMPLIES₂(not₁(x), not₁(y)) -> IMPLIES₂(y, and₂(x, y))

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

POL(IMPLIES₂(x₁, x₂)) = x₁ + 2·x₂
POL(and₂(x₁, x₂)) = 1 + x₁
POL(false) = 0
POL(not₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented:

and₂(x, false) -> false
and₂(x, not₁(false)) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

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

and₂(x, false) -> false
and₂(x, not₁(false)) -> x
not₁(not₁(x)) -> x
implies₂(false, y) -> not₁(false)
implies₂(x, false) -> not₁(x)
implies₂(not₁(x), not₁(y)) -> implies₂(y, and₂(x, y))

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.