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)  =  EVENODD1(x1)
0  =  0
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[EVENODD1, 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.