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:

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(0, s(0)) → false
evenodd(s(x), s(0)) → evenodd(x, 0)

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(0, s(0)) → false
evenodd(s(x), s(0)) → evenodd(x, 0)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(0, s(0)) → false
evenodd(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

not(true)
not(false)
evenodd(x0, 0)
evenodd(0, s(0))
evenodd(s(x0), s(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:

EVENODD(x, 0) → EVENODD(x, s(0))
EVENODD(s(x), s(0)) → EVENODD(x, 0)
EVENODD(x, 0) → NOT(evenodd(x, s(0)))

The TRS R consists of the following rules:

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(0, s(0)) → false
evenodd(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

not(true)
not(false)
evenodd(x0, 0)
evenodd(0, s(0))
evenodd(s(x0), s(0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

EVENODD(x, 0) → EVENODD(x, s(0))
EVENODD(s(x), s(0)) → EVENODD(x, 0)
EVENODD(x, 0) → NOT(evenodd(x, s(0)))

The TRS R consists of the following rules:

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(0, s(0)) → false
evenodd(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

not(true)
not(false)
evenodd(x0, 0)
evenodd(0, s(0))
evenodd(s(x0), s(0))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

EVENODD(x, 0) → EVENODD(x, s(0))
EVENODD(s(x), s(0)) → EVENODD(x, 0)
EVENODD(x, 0) → NOT(evenodd(x, s(0)))

The TRS R consists of the following rules:

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(0, s(0)) → false
evenodd(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

not(true)
not(false)
evenodd(x0, 0)
evenodd(0, s(0))
evenodd(s(x0), s(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

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

EVENODD(x, 0) → EVENODD(x, s(0))
EVENODD(s(x), s(0)) → EVENODD(x, 0)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(0, s(0)) → false
evenodd(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

not(true)
not(false)
evenodd(x0, 0)
evenodd(0, s(0))
evenodd(s(x0), s(0))

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.


EVENODD(s(x), s(0)) → EVENODD(x, 0)
The remaining pairs can at least be oriented weakly.

EVENODD(x, 0) → EVENODD(x, s(0))
Used ordering: Combined order from the following AFS and order.
EVENODD(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ DependencyGraphProof

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

EVENODD(x, 0) → EVENODD(x, s(0))

The TRS R consists of the following rules:

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(0, s(0)) → false
evenodd(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

not(true)
not(false)
evenodd(x0, 0)
evenodd(0, s(0))
evenodd(s(x0), s(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.