Termination w.r.t. Q proof of /tmp/tmp83mBBs/8.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:

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

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, y) → cond(and(gr(x, 0), gr(y, 0)), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
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, x1)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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, y) → AND(gr(x, 0), gr(y, 0))
COND(true, x, y) → P(x)
COND(true, x, y) → GR(x, 0)
COND(true, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND(true, x, y) → COND(and(gr(x, 0), gr(y, 0)), p(x), p(y))
COND(true, x, y) → P(y)

The TRS R consists of the following rules:

cond(true, x, y) → cond(and(gr(x, 0), gr(y, 0)), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
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, x1)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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, y) → AND(gr(x, 0), gr(y, 0))
COND(true, x, y) → P(x)
COND(true, x, y) → GR(x, 0)
COND(true, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND(true, x, y) → COND(and(gr(x, 0), gr(y, 0)), p(x), p(y))
COND(true, x, y) → P(y)

The TRS R consists of the following rules:

cond(true, x, y) → cond(and(gr(x, 0), gr(y, 0)), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
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, x1)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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:

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x, y) → cond(and(gr(x, 0), gr(y, 0)), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
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, x1)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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:

GR(s(x), s(y)) → GR(x, y)

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

cond(true, x0, x1)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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, x1)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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:

GR(s(x), s(y)) → GR(x, y)

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:

GR(s(x), s(y)) → GR(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ 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, y) → COND(and(gr(x, 0), gr(y, 0)), p(x), p(y))

The TRS R consists of the following rules:

cond(true, x, y) → cond(and(gr(x, 0), gr(y, 0)), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
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, x1)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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, y) → COND(and(gr(x, 0), gr(y, 0)), p(x), p(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond(true, x0, x1)
and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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, x1)

↳ 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, y) → COND(and(gr(x, 0), gr(y, 0)), p(x), p(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

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

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1))
COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

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

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1))
COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1))
COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), p(s(x0)), p(y1)) at position [1] we obtained the following new rules:

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

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

COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0))
COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, y0, 0) → COND(and(gr(y0, 0), false), p(y0), p(0)) at position [0] we obtained the following new rules:

COND(true, y0, 0) → COND(false, p(y0), p(0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

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

COND(true, y0, 0) → COND(false, p(y0), p(0))
COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

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

COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, 0, y1) → COND(and(false, gr(y1, 0)), p(0), p(y1)) at position [0] we obtained the following new rules:

COND(true, 0, y1) → COND(false, p(0), p(y1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

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

COND(true, 0, y1) → COND(false, p(0), p(y1))
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

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

COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), p(s(x0))) at position [2] we obtained the following new rules:

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND(true, s(x0), y1) → COND(and(true, gr(y1, 0)), x0, p(y1)) at position [0] we obtained the following new rules:

COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))
COND(true, s(y0), 0) → COND(and(true, false), y0, p(0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

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

COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), 0) → COND(and(true, false), y0, p(0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

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

COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

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

COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0)))
COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, s(y0), s(x0)) → COND(and(true, true), y0, p(s(x0))) at position [0] we obtained the following new rules:

COND(true, s(y0), s(x0)) → COND(true, y0, p(s(x0)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), s(x0)) → COND(true, y0, p(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, s(y0), s(x0)) → COND(true, y0, p(s(x0))) at position [2] we obtained the following new rules:

COND(true, s(y0), s(x0)) → COND(true, y0, x0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

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

COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0)
COND(true, s(y0), s(x0)) → COND(true, y0, x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND(true, y0, s(x0)) → COND(and(gr(y0, 0), true), p(y0), x0) at position [0] we obtained the following new rules:

COND(true, 0, s(y1)) → COND(and(false, true), p(0), y1)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, 0, s(y1)) → COND(and(false, true), p(0), y1)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ UsableRulesProof

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ UsableRulesProof

                                                                                  ↳ QDP

                                                                                    ↳ QReductionProof

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, x0)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
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(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ UsableRulesProof

                                                                                  ↳ QDP

                                                                                    ↳ QReductionProof

                                                                                      ↳ QDP

                                                                                        ↳ Rewriting

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1)

The TRS R consists of the following rules:

and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, s(x0), s(y1)) → COND(and(true, true), p(s(x0)), y1) at position [0] we obtained the following new rules:

COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ UsableRulesProof

                                                                                  ↳ QDP

                                                                                    ↳ QReductionProof

                                                                                      ↳ QDP

                                                                                        ↳ Rewriting

                                                                                          ↳ QDP

                                                                                            ↳ UsableRulesProof

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

COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)
COND(true, s(y0), s(x0)) → COND(true, y0, x0)

The TRS R consists of the following rules:

and(true, true) → true
p(s(x)) → x

The set Q consists of the following terms:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ UsableRulesProof

                                                                                  ↳ QDP

                                                                                    ↳ QReductionProof

                                                                                      ↳ QDP

                                                                                        ↳ Rewriting

                                                                                          ↳ QDP

                                                                                            ↳ UsableRulesProof

                                                                                              ↳ QDP

                                                                                                ↳ QReductionProof

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

COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)
COND(true, s(y0), s(x0)) → COND(true, y0, x0)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

and(true, true)
and(x0, false)
and(false, 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.

and(true, true)
and(x0, false)
and(false, x0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ UsableRulesProof

                                                                                  ↳ QDP

                                                                                    ↳ QReductionProof

                                                                                      ↳ QDP

                                                                                        ↳ Rewriting

                                                                                          ↳ QDP

                                                                                            ↳ UsableRulesProof

                                                                                              ↳ QDP

                                                                                                ↳ QReductionProof

                                                                                                  ↳ QDP

                                                                                                    ↳ Rewriting

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

COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1)
COND(true, s(y0), s(x0)) → COND(true, y0, x0)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND(true, s(x0), s(y1)) → COND(true, p(s(x0)), y1) at position [1] we obtained the following new rules:

COND(true, s(x0), s(y1)) → COND(true, x0, y1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ UsableRulesProof

                                                                                  ↳ QDP

                                                                                    ↳ QReductionProof

                                                                                      ↳ QDP

                                                                                        ↳ Rewriting

                                                                                          ↳ QDP

                                                                                            ↳ UsableRulesProof

                                                                                              ↳ QDP

                                                                                                ↳ QReductionProof

                                                                                                  ↳ QDP

                                                                                                    ↳ Rewriting

                                                                                                      ↳ QDP

                                                                                                        ↳ UsableRulesProof

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ UsableRulesProof

                                                                                  ↳ QDP

                                                                                    ↳ QReductionProof

                                                                                      ↳ 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(y0), s(x0)) → COND(true, y0, x0)

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

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

                            ↳ Rewriting

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Rewriting

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Narrowing

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ QDP

                                                                                ↳ UsableRulesProof

                                                                                  ↳ QDP

                                                                                    ↳ QReductionProof

                                                                                      ↳ QDP

                                                                                        ↳ Rewriting

                                                                                          ↳ QDP

                                                                                            ↳ UsableRulesProof

                                                                                              ↳ QDP

                                                                                                ↳ QReductionProof

                                                                                                  ↳ QDP

                                                                                                    ↳ Rewriting

                                                                                                      ↳ QDP

                                                                                                        ↳ UsableRulesProof

                                                                                                          ↳ QDP

                                                                                                            ↳ QReductionProof

                                                                                                              ↳ QDP

                                                                                                                ↳ QDPSizeChangeProof

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)

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

COND(true, s(y0), s(x0)) → COND(true, y0, x0)
The graph contains the following edges 1 >= 1, 2 > 2, 3 > 3