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.