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
  ↳ 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)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


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: Polynomial interpretation [25,35]:

POL(EVENODD(x1, x2)) = (1/2)x_1   
POL(s(x1)) = 3/4 + x_1   
POL(0) = 0   
The value of delta used in the strict ordering is 3/8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.