Termination w.r.t. Q proof of /tmp/tmpsgH7

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:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))

The signature Sigma is {cond}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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

COND(true, x) → P(x)
COND(true, x) → EVEN(x)
EVEN(s(s(x))) → EVEN(x)
COND(true, x) → GR(x, 0)
COND(true, x) → AND(even(x), gr(x, 0))
COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

COND(true, x) → P(x)
COND(true, x) → EVEN(x)
EVEN(s(s(x))) → EVEN(x)
COND(true, x) → GR(x, 0)
COND(true, x) → AND(even(x), gr(x, 0))
COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 5 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

EVEN(s(s(x))) → EVEN(x)

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

EVEN(s(s(x))) → EVEN(x)

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

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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.

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

EVEN(s(s(x))) → EVEN(x)

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:

EVEN(s(s(x))) → EVEN(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

cond(true, x) → cond(and(even(x), gr(x, 0)), p(x))
and(x, false) → false
and(false, x) → false
and(true, true) → true
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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.

cond(true, x0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

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

COND(true, x) → COND(and(even(x), gr(x, 0)), p(x))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND(true, x) → COND(and(even(x), gr(x, 0)), p(x)) at position [0] we obtained the following new rules:

COND(true, 0) → COND(and(even(0), false), p(0))
COND(true, s(0)) → COND(and(false, gr(s(0), 0)), p(s(0)))
COND(true, 0) → COND(and(true, gr(0, 0)), p(0))
COND(true, s(x0)) → COND(and(even(s(x0)), true), p(s(x0)))
COND(true, s(s(x0))) → COND(and(even(x0), gr(s(s(x0)), 0)), p(s(s(x0))))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

COND(true, s(0)) → COND(and(false, gr(s(0), 0)), p(s(0)))
COND(true, s(x0)) → COND(and(even(s(x0)), true), p(s(x0)))
COND(true, 0) → COND(and(true, gr(0, 0)), p(0))
COND(true, 0) → COND(and(even(0), false), p(0))
COND(true, s(s(x0))) → COND(and(even(x0), gr(s(s(x0)), 0)), p(s(s(x0))))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                ↳ QDP

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

COND(true, 0) → COND(and(true, gr(0, 0)), p(0))
COND(true, 0) → COND(and(even(0), false), p(0))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                ↳ QDP

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

COND(true, 0) → COND(and(true, gr(0, 0)), p(0))
COND(true, 0) → COND(and(even(0), false), p(0))

The TRS R consists of the following rules:

even(0) → true
and(x, false) → false
and(false, x) → false
p(0) → 0
gr(0, x) → false
and(true, true) → true

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, 0) → COND(and(even(0), false), p(0)) at position [0] we obtained the following new rules:

COND(true, 0) → COND(false, p(0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                ↳ QDP

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

COND(true, 0) → COND(false, p(0))
COND(true, 0) → COND(and(true, gr(0, 0)), p(0))

The TRS R consists of the following rules:

even(0) → true
and(x, false) → false
and(false, x) → false
p(0) → 0
gr(0, x) → false
and(true, true) → true

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ UsableRulesProof

                                ↳ QDP

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

COND(true, 0) → COND(and(true, gr(0, 0)), p(0))

The TRS R consists of the following rules:

even(0) → true
and(x, false) → false
and(false, x) → false
p(0) → 0
gr(0, x) → false
and(true, true) → true

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ UsableRulesProof

                                                ↳ QDP

                                                  ↳ QReductionProof

                                ↳ QDP

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

COND(true, 0) → COND(and(true, gr(0, 0)), p(0))

The TRS R consists of the following rules:

gr(0, x) → false
and(x, false) → false
and(true, true) → true
p(0) → 0

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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.

even(0)
even(s(0))
even(s(s(x0)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ UsableRulesProof

                                                ↳ QDP

                                                  ↳ QReductionProof

                                                    ↳ QDP

                                                      ↳ Rewriting

                                ↳ QDP

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

COND(true, 0) → COND(and(true, gr(0, 0)), p(0))

The TRS R consists of the following rules:

gr(0, x) → false
and(x, false) → false
and(true, true) → true
p(0) → 0

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, 0) → COND(and(true, gr(0, 0)), p(0)) at position [0,1] we obtained the following new rules:

COND(true, 0) → COND(and(true, false), p(0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ UsableRulesProof

                                                ↳ QDP

                                                  ↳ QReductionProof

                                                    ↳ QDP

                                                      ↳ Rewriting

                                                        ↳ QDP

                                                          ↳ DependencyGraphProof

                                ↳ QDP

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

COND(true, 0) → COND(and(true, false), p(0))

The TRS R consists of the following rules:

gr(0, x) → false
and(x, false) → false
and(true, true) → true
p(0) → 0

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

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

COND(true, s(0)) → COND(and(false, gr(s(0), 0)), p(s(0)))
COND(true, s(x0)) → COND(and(even(s(x0)), true), p(s(x0)))
COND(true, s(s(x0))) → COND(and(even(x0), gr(s(s(x0)), 0)), p(s(s(x0))))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(0, x) → false
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

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

COND(true, s(0)) → COND(and(false, gr(s(0), 0)), p(s(0)))
COND(true, s(x0)) → COND(and(even(s(x0)), true), p(s(x0)))
COND(true, s(s(x0))) → COND(and(even(x0), gr(s(s(x0)), 0)), p(s(s(x0))))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, s(0)) → COND(and(false, gr(s(0), 0)), p(s(0))) at position [0] we obtained the following new rules:

COND(true, s(0)) → COND(false, p(s(0)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

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

COND(true, s(x0)) → COND(and(even(s(x0)), true), p(s(x0)))
COND(true, s(0)) → COND(false, p(s(0)))
COND(true, s(s(x0))) → COND(and(even(x0), gr(s(s(x0)), 0)), p(s(s(x0))))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

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

COND(true, s(x0)) → COND(and(even(s(x0)), true), p(s(x0)))
COND(true, s(s(x0))) → COND(and(even(x0), gr(s(s(x0)), 0)), p(s(s(x0))))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, s(x0)) → COND(and(even(s(x0)), true), p(s(x0))) at position [1] we obtained the following new rules:

COND(true, s(x0)) → COND(and(even(s(x0)), true), x0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

                                                ↳ QDP

                                                  ↳ Rewriting

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

COND(true, s(x0)) → COND(and(even(s(x0)), true), x0)
COND(true, s(s(x0))) → COND(and(even(x0), gr(s(s(x0)), 0)), p(s(s(x0))))

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, s(s(x0))) → COND(and(even(x0), gr(s(s(x0)), 0)), p(s(s(x0)))) at position [0,1] we obtained the following new rules:

COND(true, s(s(x0))) → COND(and(even(x0), true), p(s(s(x0))))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

                                                ↳ QDP

                                                  ↳ Rewriting

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

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

COND(true, s(s(x0))) → COND(and(even(x0), true), p(s(s(x0))))
COND(true, s(x0)) → COND(and(even(s(x0)), true), x0)

The TRS R consists of the following rules:

even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
gr(s(x), 0) → true
and(x, false) → false
and(false, x) → false
and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

                                                ↳ QDP

                                                  ↳ Rewriting

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

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

COND(true, s(s(x0))) → COND(and(even(x0), true), p(s(s(x0))))
COND(true, s(x0)) → COND(and(even(s(x0)), true), x0)

The TRS R consists of the following rules:

even(s(0)) → false
even(s(s(x))) → even(x)
and(false, x) → false
and(true, true) → true
even(0) → true
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))
p(0)
p(s(x0))

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.

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(y))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

                                                ↳ QDP

                                                  ↳ Rewriting

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

                                                            ↳ QDP

                                                              ↳ Rewriting

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

COND(true, s(s(x0))) → COND(and(even(x0), true), p(s(s(x0))))
COND(true, s(x0)) → COND(and(even(s(x0)), true), x0)

The TRS R consists of the following rules:

even(s(0)) → false
even(s(s(x))) → even(x)
and(false, x) → false
and(true, true) → true
even(0) → true
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, s(s(x0))) → COND(and(even(x0), true), p(s(s(x0)))) at position [1] we obtained the following new rules:

COND(true, s(s(x0))) → COND(and(even(x0), true), s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

                                                ↳ QDP

                                                  ↳ Rewriting

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ UsableRulesProof

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

COND(true, s(s(x0))) → COND(and(even(x0), true), s(x0))
COND(true, s(x0)) → COND(and(even(s(x0)), true), x0)

The TRS R consists of the following rules:

even(s(0)) → false
even(s(s(x))) → even(x)
and(false, x) → false
and(true, true) → true
even(0) → true
p(s(x)) → x

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

                                                ↳ QDP

                                                  ↳ Rewriting

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ UsableRulesProof

                                                                    ↳ QDP

                                                                      ↳ QReductionProof

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

COND(true, s(s(x0))) → COND(and(even(x0), true), s(x0))
COND(true, s(x0)) → COND(and(even(s(x0)), true), x0)

The TRS R consists of the following rules:

even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true
and(false, x) → false
and(true, true) → true

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))
p(0)
p(s(x0))

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.

p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

                                                ↳ QDP

                                                  ↳ Rewriting

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ UsableRulesProof

                                                                    ↳ QDP

                                                                      ↳ QReductionProof

                                                                        ↳ QDP

                                                                          ↳ Narrowing

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

COND(true, s(s(x0))) → COND(and(even(x0), true), s(x0))
COND(true, s(x0)) → COND(and(even(s(x0)), true), x0)

The TRS R consists of the following rules:

even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true
and(false, x) → false
and(true, true) → true

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND(true, s(x0)) → COND(and(even(s(x0)), true), x0) at position [0] we obtained the following new rules:

COND(true, s(0)) → COND(and(false, true), 0)
COND(true, s(s(x0))) → COND(and(even(x0), true), s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

                                                ↳ QDP

                                                  ↳ Rewriting

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ UsableRulesProof

                                                                    ↳ QDP

                                                                      ↳ QReductionProof

                                                                        ↳ QDP

                                                                          ↳ Narrowing

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

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

COND(true, s(0)) → COND(and(false, true), 0)
COND(true, s(s(x0))) → COND(and(even(x0), true), s(x0))

The TRS R consists of the following rules:

even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true
and(false, x) → false
and(true, true) → true

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))

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

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

                                                ↳ QDP

                                                  ↳ Rewriting

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ UsableRulesProof

                                                                    ↳ QDP

                                                                      ↳ QReductionProof

                                                                        ↳ QDP

                                                                          ↳ Narrowing

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

                                                                                ↳ QDP

                                                                                  ↳ ForwardInstantiation

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

COND(true, s(s(x0))) → COND(and(even(x0), true), s(x0))

The TRS R consists of the following rules:

even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true
and(false, x) → false
and(true, true) → true

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule COND(true, s(s(x0))) → COND(and(even(x0), true), s(x0)) we obtained the following new rules:

COND(true, s(s(s(y_1)))) → COND(and(even(s(y_1)), true), s(s(y_1)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ AND

                                ↳ QDP

                                ↳ QDP

                                  ↳ UsableRulesProof

                                    ↳ QDP

                                      ↳ Rewriting

                                        ↳ QDP

                                          ↳ DependencyGraphProof

                                            ↳ QDP

                                              ↳ Rewriting

                                                ↳ QDP

                                                  ↳ Rewriting

                                                    ↳ QDP

                                                      ↳ UsableRulesProof

                                                        ↳ QDP

                                                          ↳ QReductionProof

                                                            ↳ QDP

                                                              ↳ Rewriting

                                                                ↳ QDP

                                                                  ↳ UsableRulesProof

                                                                    ↳ QDP

                                                                      ↳ QReductionProof

                                                                        ↳ QDP

                                                                          ↳ Narrowing

                                                                            ↳ QDP

                                                                              ↳ DependencyGraphProof

                                                                                ↳ QDP

                                                                                  ↳ ForwardInstantiation

                                                                                    ↳ QDP

                                                                                      ↳ QDPSizeChangeProof

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

COND(true, s(s(s(y_1)))) → COND(and(even(s(y_1)), true), s(s(y_1)))

The TRS R consists of the following rules:

even(s(0)) → false
even(s(s(x))) → even(x)
even(0) → true
and(false, x) → false
and(true, true) → true

The set Q consists of the following terms:

and(x0, false)
and(false, x0)
and(true, true)
even(0)
even(s(0))
even(s(s(x0)))

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:

COND(true, s(s(s(y_1)))) → COND(and(even(s(y_1)), true), s(s(y_1)))
The graph contains the following edges 2 > 2