Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/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))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

EVENODD₂(x, 0) -> EVENODD₂(x, s₁(0))

Used ordering: Combined order from the following AFS and order.
EVENODD₂(x1, x2) = x1
0 = 0
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

0 > s₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ 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 [13,14,18] contains 0 SCCs with 1 less node.