Termination w.r.t. Q proof of /tmp/tmp6m-jo6/#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:

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

  ↳ RRRPoloQTRSProof

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 following Q TRS is given: 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 following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

evenodd(0, s(0)) → false

Used ordering:
Polynomial interpretation [25]:

POL(0) = 0
POL(evenodd(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(false) = 0
POL(not(x₁)) = x₁
POL(s(x₁)) = x₁
POL(true) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ 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(s(x), s(0)) → evenodd(x, 0)

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ 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(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

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

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(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ 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(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

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

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

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

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(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

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)

R is empty.
The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ Overlay + Local Confluence

        ↳ QTRS

          ↳ DependencyPairsProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ UsableRulesProof

                    ↳ QDP

                      ↳ QReductionProof

                        ↳ QDP

                          ↳ QDPSizeChangeProof

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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

EVENODD(s(x), s(0)) → EVENODD(x, 0)
The graph contains the following edges 1 > 1, 2 > 2
EVENODD(x, 0) → EVENODD(x, s(0))
The graph contains the following edges 1 >= 1