Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/AG01SLASHHASH3.37.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:

not₁(true) -> false
not₁(false) -> true
evenodd₂(x, 0) -> not₁(evenodd₂(x, s₁(0)))
evenodd₂(0, s₁(0)) -> false
evenodd₂(s₁(x), s₁(0)) -> evenodd₂(x, 0)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

not₁(true) -> false
not₁(false) -> true
evenodd₂(x, 0) -> not₁(evenodd₂(x, s₁(0)))
evenodd₂(0, s₁(0)) -> false
evenodd₂(s₁(x), s₁(0)) -> evenodd₂(x, 0)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

not₁(true) -> false
not₁(false) -> true
evenodd₂(x, 0) -> not₁(evenodd₂(x, s₁(0)))
evenodd₂(0, s₁(0)) -> false
evenodd₂(s₁(x), s₁(0)) -> evenodd₂(x, 0)

The set Q consists of the following terms:

not₁(true)
not₁(false)
evenodd₂(x0, 0)
evenodd₂(0, s₁(0))
evenodd₂(s₁(x0), s₁(0))

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

EVENODD₂(x, 0) -> EVENODD₂(x, s₁(0))
EVENODD₂(s₁(x), s₁(0)) -> EVENODD₂(x, 0)
EVENODD₂(x, 0) -> NOT₁(evenodd₂(x, s₁(0)))

The TRS R consists of the following rules:

not₁(true) -> false
not₁(false) -> true
evenodd₂(x, 0) -> not₁(evenodd₂(x, s₁(0)))
evenodd₂(0, s₁(0)) -> false
evenodd₂(s₁(x), s₁(0)) -> evenodd₂(x, 0)

The set Q consists of the following terms:

not₁(true)
not₁(false)
evenodd₂(x0, 0)
evenodd₂(0, s₁(0))
evenodd₂(s₁(x0), s₁(0))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

EVENODD₂(x, 0) -> EVENODD₂(x, s₁(0))
EVENODD₂(s₁(x), s₁(0)) -> EVENODD₂(x, 0)
EVENODD₂(x, 0) -> NOT₁(evenodd₂(x, s₁(0)))

The TRS R consists of the following rules:

not₁(true) -> false
not₁(false) -> true
evenodd₂(x, 0) -> not₁(evenodd₂(x, s₁(0)))
evenodd₂(0, s₁(0)) -> false
evenodd₂(s₁(x), s₁(0)) -> evenodd₂(x, 0)

The set Q consists of the following terms:

not₁(true)
not₁(false)
evenodd₂(x0, 0)
evenodd₂(0, s₁(0))
evenodd₂(s₁(x0), s₁(0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

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

EVENODD₂(x, 0) -> EVENODD₂(x, s₁(0))
EVENODD₂(s₁(x), s₁(0)) -> EVENODD₂(x, 0)

The TRS R consists of the following rules:

not₁(true) -> false
not₁(false) -> true
evenodd₂(x, 0) -> not₁(evenodd₂(x, s₁(0)))
evenodd₂(0, s₁(0)) -> false
evenodd₂(s₁(x), s₁(0)) -> evenodd₂(x, 0)

The set Q consists of the following terms:

not₁(true)
not₁(false)
evenodd₂(x0, 0)
evenodd₂(0, s₁(0))
evenodd₂(s₁(x0), s₁(0))

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.

EVENODD₂(s₁(x), s₁(0)) -> EVENODD₂(x, 0)

Used argument filtering: EVENODD₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

EVENODD₂(x, 0) -> EVENODD₂(x, s₁(0))

The TRS R consists of the following rules:

not₁(true) -> false
not₁(false) -> true
evenodd₂(x, 0) -> not₁(evenodd₂(x, s₁(0)))
evenodd₂(0, s₁(0)) -> false
evenodd₂(s₁(x), s₁(0)) -> evenodd₂(x, 0)

The set Q consists of the following terms:

not₁(true)
not₁(false)
evenodd₂(x0, 0)
evenodd₂(0, s₁(0))
evenodd₂(s₁(x0), s₁(0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.