Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/SK90SLASH2.35.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:

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

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.