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

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

The TRS R 2 is

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

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(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(s(x), y) → GR(s(x), y)
COND2(false, x, y) → P(x)
COND2(true, x, y) → COND1(y, y)
GR(s(x), s(y)) → GR(x, y)
NEQ(s(x), s(y)) → NEQ(x, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(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(s(x), y) → GR(s(x), y)
COND2(false, x, y) → P(x)
COND2(true, x, y) → COND1(y, y)
GR(s(x), s(y)) → GR(x, y)
NEQ(s(x), s(y)) → NEQ(x, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(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 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

NEQ(s(x), s(y)) → NEQ(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(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

              ↳ QDP

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

NEQ(s(x), s(y)) → NEQ(x, y)

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

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(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(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

NEQ(s(x), s(y)) → NEQ(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:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

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

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(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(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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)

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

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(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(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                ↳ UsableRulesProof

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

COND2(false, 0, y1) → COND1(0, y1)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)

The TRS R consists of the following rules:

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

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

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                ↳ UsableRulesProof

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

COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)

The TRS R consists of the following rules:

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

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

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                ↳ UsableRulesProof

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

COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)

The TRS R consists of the following rules:

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

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.

p(0)
p(s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonInfProof

                ↳ UsableRulesProof

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

COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, s(x0), y1) → COND1(x0, y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus [18] with the following steps:
Note that final constraints are written in bold face.

For Pair COND2(true, x, y) → COND1(y, y) the following chains were created:

We consider the chain COND1(s(x), y) → COND2(gr(s(x), y), s(x), y), COND2(true, x, y) → COND1(y, y) which results in the following constraint:
(1)    (COND2(gr(s(x4), x5), s(x4), x5)=COND2(true, x6, x7) ⇒ COND2(true, x6, x7)≥COND1(x7, x7))

We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
(2)    (s(x4)=x26∧gr(x26, x5)=true ⇒ COND2(true, s(x4), x5)≥COND1(x5, x5))

We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x26, x5)=true which results in the following new constraints:
(3)    (true=true∧s(x4)=s(x27) ⇒ COND2(true, s(x4), 0)≥COND1(0, 0))

(4)    (gr(x28, x29)=true∧s(x4)=s(x28)∧(∀x30:gr(x28, x29)=true∧s(x30)=x28 ⇒ COND2(true, s(x30), x29)≥COND1(x29, x29)) ⇒ COND2(true, s(x4), s(x29))≥COND1(s(x29), s(x29)))

We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:
(5)    (COND2(true, s(x4), 0)≥COND1(0, 0))

We simplified constraint (4) using rules (I), (II), (III), (IV) which results in the following new constraint:
(6)    (gr(x28, x29)=true ⇒ COND2(true, s(x28), s(x29))≥COND1(s(x29), s(x29)))

We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on gr(x28, x29)=true which results in the following new constraints:
(7)    (true=true ⇒ COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))

(8)    (gr(x33, x34)=true∧(gr(x33, x34)=true ⇒ COND2(true, s(x33), s(x34))≥COND1(s(x34), s(x34))) ⇒ COND2(true, s(s(x33)), s(s(x34)))≥COND1(s(s(x34)), s(s(x34))))

We simplified constraint (7) using rules (I), (II) which results in the following new constraint:
(9)    (COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))

We simplified constraint (8) using rule (VI) where we applied the induction hypothesis (gr(x33, x34)=true ⇒ COND2(true, s(x33), s(x34))≥COND1(s(x34), s(x34))) with σ = [ ] which results in the following new constraint:
(10)    (COND2(true, s(x33), s(x34))≥COND1(s(x34), s(x34)) ⇒ COND2(true, s(s(x33)), s(s(x34)))≥COND1(s(s(x34)), s(s(x34))))

For Pair COND2(false, s(x0), y1) → COND1(x0, y1) the following chains were created:

We consider the chain COND1(s(x), y) → COND2(gr(s(x), y), s(x), y), COND2(false, s(x0), y1) → COND1(x0, y1) which results in the following constraint:
(11)    (COND2(gr(s(x12), x13), s(x12), x13)=COND2(false, s(x14), x15) ⇒ COND2(false, s(x14), x15)≥COND1(x14, x15))

We simplified constraint (11) using rules (I), (II), (III), (VII) which results in the following new constraint:
(12)    (s(x12)=x36∧gr(x36, x13)=false ⇒ COND2(false, s(x12), x13)≥COND1(x12, x13))

We simplified constraint (12) using rule (V) (with possible (I) afterwards) using induction on gr(x36, x13)=false which results in the following new constraints:
(13)    (gr(x38, x39)=false∧s(x12)=s(x38)∧(∀x40:gr(x38, x39)=false∧s(x40)=x38 ⇒ COND2(false, s(x40), x39)≥COND1(x40, x39)) ⇒ COND2(false, s(x12), s(x39))≥COND1(x12, s(x39)))

(14)    (false=false∧s(x12)=0 ⇒ COND2(false, s(x12), x41)≥COND1(x12, x41))

We simplified constraint (13) using rules (I), (II), (III), (IV) which results in the following new constraint:
(15)    (gr(x38, x39)=false ⇒ COND2(false, s(x38), s(x39))≥COND1(x38, s(x39)))

We solved constraint (14) using rules (I), (II).We simplified constraint (15) using rule (V) (with possible (I) afterwards) using induction on gr(x38, x39)=false which results in the following new constraints:
(16)    (gr(x43, x44)=false∧(gr(x43, x44)=false ⇒ COND2(false, s(x43), s(x44))≥COND1(x43, s(x44))) ⇒ COND2(false, s(s(x43)), s(s(x44)))≥COND1(s(x43), s(s(x44))))

(17)    (false=false ⇒ COND2(false, s(0), s(x45))≥COND1(0, s(x45)))

We simplified constraint (16) using rule (VI) where we applied the induction hypothesis (gr(x43, x44)=false ⇒ COND2(false, s(x43), s(x44))≥COND1(x43, s(x44))) with σ = [ ] which results in the following new constraint:
(18)    (COND2(false, s(x43), s(x44))≥COND1(x43, s(x44)) ⇒ COND2(false, s(s(x43)), s(s(x44)))≥COND1(s(x43), s(s(x44))))

We simplified constraint (17) using rules (I), (II) which results in the following new constraint:
(19)    (COND2(false, s(0), s(x45))≥COND1(0, s(x45)))

For Pair COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) the following chains were created:

We consider the chain COND2(true, x, y) → COND1(y, y), COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) which results in the following constraint:
(20) (COND1(x17, x17)=COND1(s(x18), x19) ⇒ COND1(s(x18), x19)≥COND2(gr(s(x18), x19), s(x18), x19))

We simplified constraint (20) using rules (I), (II), (III) which results in the following new constraint:
(21) (COND1(s(x18), s(x18))≥COND2(gr(s(x18), s(x18)), s(x18), s(x18)))
We consider the chain COND2(false, s(x0), y1) → COND1(x0, y1), COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) which results in the following constraint:
(22) (COND1(x20, x21)=COND1(s(x22), x23) ⇒ COND1(s(x22), x23)≥COND2(gr(s(x22), x23), s(x22), x23))

We simplified constraint (22) using rules (I), (II), (III) which results in the following new constraint:
(23) (COND1(s(x22), x21)≥COND2(gr(s(x22), x21), s(x22), x21))

To summarize, we get the following constraints P_≥ for the following pairs.

COND2(true, x, y) → COND1(y, y)
- (COND2(true, s(x4), 0)≥COND1(0, 0))
- (COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))
- (COND2(true, s(x33), s(x34))≥COND1(s(x34), s(x34)) ⇒ COND2(true, s(s(x33)), s(s(x34)))≥COND1(s(s(x34)), s(s(x34))))
COND2(false, s(x0), y1) → COND1(x0, y1)
- (COND2(false, s(x43), s(x44))≥COND1(x43, s(x44)) ⇒ COND2(false, s(s(x43)), s(s(x44)))≥COND1(s(x43), s(s(x44))))
- (COND2(false, s(0), s(x45))≥COND1(0, s(x45)))
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
- (COND1(s(x18), s(x18))≥COND2(gr(s(x18), s(x18)), s(x18), s(x18)))
- (COND1(s(x22), x21)≥COND2(gr(s(x22), x21), s(x22), x21))

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [18]:

POL(0) = 1
POL(COND1(x₁, x₂)) = -1 + x₁
POL(COND2(x₁, x₂, x₃)) = -1 - x₁ + x₂
POL(c) = -1
POL(false) = 0
POL(gr(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

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

The following pairs are in P_bound:

COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)

The following rules are usable:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonInfProof

                                          ↳ QDP

                                            ↳ Instantiation

                ↳ UsableRulesProof

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

COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) we obtained the following new rules:

COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonInfProof

                                          ↳ QDP

                                            ↳ Instantiation

                                              ↳ QDP

                                                ↳ Rewriting

                ↳ UsableRulesProof

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

COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0))
COND2(true, x, y) → COND1(y, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0)) at position [0] we obtained the following new rules:

COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonInfProof

                                          ↳ QDP

                                            ↳ Instantiation

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Instantiation

                ↳ UsableRulesProof

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

COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
COND2(true, x, y) → COND1(y, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

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

COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonInfProof

                                          ↳ QDP

                                            ↳ Instantiation

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                ↳ UsableRulesProof

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

COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0)) at position [0] we obtained the following new rules:

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonInfProof

                                          ↳ QDP

                                            ↳ Instantiation

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                ↳ UsableRulesProof

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

COND1(s(s(x0)), s(s(x0))) → COND2(gr(x0, x0), s(s(x0)), s(s(x0)))
COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))
COND1(s(0), s(0)) → COND2(false, s(0), s(0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

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

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonInfProof

                                          ↳ QDP

                                            ↳ Instantiation

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Instantiation

                ↳ UsableRulesProof

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

COND1(s(s(x0)), s(s(x0))) → COND2(gr(x0, x0), s(s(x0)), s(s(x0)))
COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0)) we obtained the following new rules:

COND2(true, s(s(z0)), s(s(z0))) → COND1(s(s(z0)), s(s(z0)))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonInfProof

                                          ↳ QDP

                                            ↳ Instantiation

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Instantiation

                                                                  ↳ QDP

                                                                    ↳ MNOCProof

                ↳ UsableRulesProof

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

COND2(true, s(s(z0)), s(s(z0))) → COND1(s(s(z0)), s(s(z0)))
COND1(s(s(x0)), s(s(x0))) → COND2(gr(x0, x0), s(s(x0)), s(s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonInfProof

                                          ↳ QDP

                                            ↳ Instantiation

                                              ↳ QDP

                                                ↳ Rewriting

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ Instantiation

                                                                  ↳ QDP

                                                                    ↳ MNOCProof

                                                                      ↳ QDP

                ↳ UsableRulesProof

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

COND2(true, s(s(z0)), s(s(z0))) → COND1(s(s(z0)), s(s(z0)))
COND1(s(s(x0)), s(s(x0))) → COND2(gr(x0, x0), s(s(x0)), s(s(x0)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all (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

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(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(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ MNOCProof

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

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

The TRS R consists of the following rules:

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

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 use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ MNOCProof

                          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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