Termination w.r.t. Q proof of TRS/AG01#3.37.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

  ↳ 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)

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:

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)

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:

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)

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 1 less node.

↳ 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)

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.

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

The remaining pairs can at least be oriented weakly.

EVENODD₂(x, 0) -> EVENODD₂(x, s₁(0))

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(EVENODD₂(x₁, x₂)) = 2·x₁
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ 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)

Q is empty.
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.