Termination w.r.t. Q proof of /tmp/tmpbNlTOQ/22.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:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
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:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
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

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

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)

The signature Sigma is {cond1, cond2}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
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:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
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:

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(true, x, y) → P(y)
COND2(false, x, y) → P(x)
COND1(true, x, y) → GR(y, 0)
COND2(true, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND2(false, x, y) → GR(x, 0)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
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:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
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:

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(true, x, y) → P(y)
COND2(false, x, y) → P(x)
COND1(true, x, y) → GR(y, 0)
COND2(true, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND2(false, x, y) → GR(x, 0)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
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:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
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:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
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:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
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:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
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.

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
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:

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
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:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
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:

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
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.

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, 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:

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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 COND1(true, x, y) → COND2(gr(y, 0), x, y) at position [0] we obtained the following new rules:

COND1(true, y0, 0) → COND2(false, y0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

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

COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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 COND2(true, x, y) → COND2(gr(y, 0), x, p(y)) at position [0] we obtained the following new rules:

COND2(true, y0, s(x0)) → COND2(true, y0, p(s(x0)))
COND2(true, y0, 0) → COND2(false, y0, p(0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

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

COND2(true, y0, 0) → COND2(false, y0, p(0))
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(true, y0, p(s(x0)))
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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 COND2(true, y0, s(x0)) → COND2(true, y0, p(s(x0))) at position [2] we obtained the following new rules:

COND2(true, y0, s(x0)) → COND2(true, y0, x0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

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

COND2(true, y0, 0) → COND2(false, y0, p(0))
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(true, y0, x0)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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 COND2(true, y0, 0) → COND2(false, y0, p(0)) at position [2] we obtained the following new rules:

COND2(true, y0, 0) → COND2(false, y0, 0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

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

COND2(true, y0, 0) → COND2(false, y0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(true, y0, x0)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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 COND2(false, x, y) → COND1(gr(x, 0), p(x), y) at position [0] we obtained the following new rules:

COND2(false, s(x0), y1) → COND1(true, p(s(x0)), y1)
COND2(false, 0, y1) → COND1(false, p(0), y1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

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

COND2(true, y0, 0) → COND2(false, y0, 0)
COND2(false, 0, y1) → COND1(false, p(0), y1)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND2(false, s(x0), y1) → COND1(true, p(s(x0)), y1)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(true, y0, x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ UsableRulesProof

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

COND2(true, y0, 0) → COND2(false, y0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND2(false, s(x0), y1) → COND1(true, p(s(x0)), y1)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(true, y0, x0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ QReductionProof

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

COND2(true, y0, 0) → COND2(false, y0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND2(false, s(x0), y1) → COND1(true, p(s(x0)), y1)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(true, y0, x0)

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

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

                            ↳ Narrowing

                              ↳ 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:

COND2(true, y0, 0) → COND2(false, y0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND2(false, s(x0), y1) → COND1(true, p(s(x0)), y1)
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(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 COND2(false, s(x0), y1) → COND1(true, p(s(x0)), y1) at position [1] we obtained the following new rules:

COND2(false, s(x0), y1) → COND1(true, x0, y1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ 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:

COND2(true, y0, 0) → COND2(false, y0, 0)
COND2(false, s(x0), y1) → COND1(true, x0, y1)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(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

                            ↳ Narrowing

                              ↳ 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:

COND2(true, y0, 0) → COND2(false, y0, 0)
COND2(false, s(x0), y1) → COND1(true, x0, y1)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(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

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ QReductionProof

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ QReductionProof

                                                                  ↳ QDP

                                                                    ↳ Instantiation

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

COND2(true, y0, 0) → COND2(false, y0, 0)
COND2(false, s(x0), y1) → COND1(true, x0, y1)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(true, y0, x0)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule COND2(false, s(x0), y1) → COND1(true, x0, y1) we obtained the following new rules:

COND2(false, s(x0), 0) → COND1(true, x0, 0)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ QReductionProof

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ QReductionProof

                                                                  ↳ QDP

                                                                    ↳ Instantiation

                                                                      ↳ QDP

                                                                        ↳ DependencyGraphProof

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

COND2(true, y0, 0) → COND2(false, y0, 0)
COND2(false, s(x0), 0) → COND1(true, x0, 0)
COND1(true, y0, s(x0)) → COND2(true, y0, s(x0))
COND1(true, y0, 0) → COND2(false, y0, 0)
COND2(true, y0, s(x0)) → COND2(true, y0, x0)

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ QReductionProof

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ QReductionProof

                                                                  ↳ QDP

                                                                    ↳ Instantiation

                                                                      ↳ QDP

                                                                        ↳ DependencyGraphProof

                                                                          ↳ AND

                                                                            ↳ QDP

                                                                              ↳ QDPSizeChangeProof

                                                                            ↳ QDP

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

COND2(false, s(x0), 0) → COND1(true, x0, 0)
COND1(true, y0, 0) → COND2(false, y0, 0)

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

COND1(true, y0, 0) → COND2(false, y0, 0)
The graph contains the following edges 2 >= 2, 3 >= 3
COND2(false, s(x0), 0) → COND1(true, x0, 0)
The graph contains the following edges 2 > 2, 3 >= 3

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ UsableRulesProof

                                                  ↳ QDP

                                                    ↳ QReductionProof

                                                      ↳ QDP

                                                        ↳ Rewriting

                                                          ↳ QDP

                                                            ↳ UsableRulesProof

                                                              ↳ QDP

                                                                ↳ QReductionProof

                                                                  ↳ QDP

                                                                    ↳ Instantiation

                                                                      ↳ QDP

                                                                        ↳ DependencyGraphProof

                                                                          ↳ AND

                                                                            ↳ QDP

                                                                            ↳ QDP

                                                                              ↳ QDPSizeChangeProof

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

COND2(true, y0, s(x0)) → COND2(true, y0, x0)

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

COND2(true, y0, s(x0)) → COND2(true, y0, x0)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3