Termination w.r.t. Q proof of /tmp/tmpGjbscn/11.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(x, y), x, y)
cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(x, 0), 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(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(x, 0), 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(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(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(x, y), x, y)
cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(x, 0), 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(true, 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:

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(x, 0), 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(true, 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 4 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(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(x, 0), 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(true, 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(true, 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(true, 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(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond1(neq(x, 0), y, y)
cond2(false, x, y) → cond1(neq(x, 0), 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(true, 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(true, 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(true, 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

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(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))
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(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

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

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

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(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))
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.
By narrowing [15] the rule COND2(false, x, y) → COND1(neq(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

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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.
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

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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.

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

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Narrowing

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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))

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

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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))

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

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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))

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ UsableRulesProof

                                                          ↳ QDP

                                                            ↳ QReductionProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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))

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.

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

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ UsableRulesProof

                                                          ↳ QDP

                                                            ↳ QReductionProof

                                                              ↳ QDP

                                                                ↳ NonInfProof

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

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

The TRS R consists of the following rules:

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

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, s(x0), y1) → COND1(true, y1, y1) the following chains were created:

We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, s(x0), y1) → COND1(true, y1, y1) which results in the following constraint:
(1)    (COND2(gr(x4, x5), x4, x5)=COND2(true, s(x6), x7) ⇒ COND2(true, s(x6), x7)≥COND1(true, x7, x7))

We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2)    (gr(x4, x5)=true∧x4=s(x6) ⇒ COND2(true, s(x6), x5)≥COND1(true, x5, x5))

We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x4, x5)=true which results in the following new constraints:
(3)    (true=true∧s(x27)=s(x6) ⇒ COND2(true, s(x6), 0)≥COND1(true, 0, 0))

(4)    (gr(x28, x29)=true∧s(x28)=s(x6)∧(∀x30:gr(x28, x29)=true∧x28=s(x30) ⇒ COND2(true, s(x30), x29)≥COND1(true, x29, x29)) ⇒ COND2(true, s(x6), s(x29))≥COND1(true, s(x29), s(x29)))

We simplified constraint (3) using rules (I), (II), (III) which results in the following new constraint:
(5)    (COND2(true, s(x27), 0)≥COND1(true, 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(true, 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(true, s(0), s(0)))

(8)    (gr(x33, x34)=true∧(gr(x33, x34)=true ⇒ COND2(true, s(x33), s(x34))≥COND1(true, s(x34), s(x34))) ⇒ COND2(true, s(s(x33)), s(s(x34)))≥COND1(true, 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(true, 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(true, s(x34), s(x34))) with σ = [ ] which results in the following new constraint:
(10)    (COND2(true, s(x33), s(x34))≥COND1(true, s(x34), s(x34)) ⇒ COND2(true, s(s(x33)), s(s(x34)))≥COND1(true, s(s(x34)), s(s(x34))))

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

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

We simplified constraint (11) using rules (I), (II), (III) which results in the following new constraint:
(12)    (gr(x12, x13)=false∧x12=s(x14) ⇒ COND2(false, s(x14), x13)≥COND1(true, x14, x13))

We simplified constraint (12) using rule (V) (with possible (I) afterwards) using induction on gr(x12, x13)=false which results in the following new constraints:
(13)    (false=false∧0=s(x14) ⇒ COND2(false, s(x14), x35)≥COND1(true, x14, x35))

(14)    (gr(x37, x38)=false∧s(x37)=s(x14)∧(∀x39:gr(x37, x38)=false∧x37=s(x39) ⇒ COND2(false, s(x39), x38)≥COND1(true, x39, x38)) ⇒ COND2(false, s(x14), s(x38))≥COND1(true, x14, s(x38)))

We solved constraint (13) using rules (I), (II).We simplified constraint (14) using rules (I), (II), (III), (IV) which results in the following new constraint:
(15)    (gr(x37, x38)=false ⇒ COND2(false, s(x37), s(x38))≥COND1(true, x37, s(x38)))

We simplified constraint (15) using rule (V) (with possible (I) afterwards) using induction on gr(x37, x38)=false which results in the following new constraints:
(16)    (false=false ⇒ COND2(false, s(0), s(x40))≥COND1(true, 0, s(x40)))

(17)    (gr(x42, x43)=false∧(gr(x42, x43)=false ⇒ COND2(false, s(x42), s(x43))≥COND1(true, x42, s(x43))) ⇒ COND2(false, s(s(x42)), s(s(x43)))≥COND1(true, s(x42), s(s(x43))))

We simplified constraint (16) using rules (I), (II) which results in the following new constraint:
(18)    (COND2(false, s(0), s(x40))≥COND1(true, 0, s(x40)))

We simplified constraint (17) using rule (VI) where we applied the induction hypothesis (gr(x42, x43)=false ⇒ COND2(false, s(x42), s(x43))≥COND1(true, x42, s(x43))) with σ = [ ] which results in the following new constraint:
(19)    (COND2(false, s(x42), s(x43))≥COND1(true, x42, s(x43)) ⇒ COND2(false, s(s(x42)), s(s(x43)))≥COND1(true, s(x42), s(s(x43))))

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

We consider the chain COND2(true, s(x0), y1) → COND1(true, y1, y1), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:
(20) (COND1(true, x17, x17)=COND1(true, x18, x19) ⇒ COND1(true, x18, x19)≥COND2(gr(x18, x19), x18, x19))

We simplified constraint (20) using rules (I), (II), (III) which results in the following new constraint:
(21) (COND1(true, x17, x17)≥COND2(gr(x17, x17), x17, x17))
We consider the chain COND2(false, s(x0), y1) → COND1(true, x0, y1), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:
(22) (COND1(true, x20, x21)=COND1(true, x22, x23) ⇒ COND1(true, x22, x23)≥COND2(gr(x22, x23), x22, x23))

We simplified constraint (22) using rules (I), (II), (III) which results in the following new constraint:
(23) (COND1(true, x20, x21)≥COND2(gr(x20, x21), x20, x21))

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

COND2(true, s(x0), y1) → COND1(true, y1, y1)
- (COND2(true, s(x27), 0)≥COND1(true, 0, 0))
- (COND2(true, s(s(x32)), s(0))≥COND1(true, s(0), s(0)))
- (COND2(true, s(x33), s(x34))≥COND1(true, s(x34), s(x34)) ⇒ COND2(true, s(s(x33)), s(s(x34)))≥COND1(true, s(s(x34)), s(s(x34))))
COND2(false, s(x0), y1) → COND1(true, x0, y1)
- (COND2(false, s(0), s(x40))≥COND1(true, 0, s(x40)))
- (COND2(false, s(x42), s(x43))≥COND1(true, x42, s(x43)) ⇒ COND2(false, s(s(x42)), s(s(x43)))≥COND1(true, s(x42), s(s(x43))))
COND1(true, x, y) → COND2(gr(x, y), x, y)
- (COND1(true, x17, x17)≥COND2(gr(x17, x17), x17, x17))
- (COND1(true, x20, x21)≥COND2(gr(x20, x21), x20, 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) = 0
POL(COND1(x₁, x₂, x₃)) = -1 - x₁ + x₂ + x₃
POL(COND2(x₁, x₂, x₃)) = -1 - x₁ + x₂ + x₃
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(true, s(x0), y1) → COND1(true, y1, y1)
COND2(false, s(x0), y1) → COND1(true, x0, y1)

The following pairs are in P_bound:

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

                                    ↳ Rewriting

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ QDP

                                                        ↳ UsableRulesProof

                                                          ↳ QDP

                                                            ↳ QReductionProof

                                                              ↳ QDP

                                                                ↳ NonInfProof

                                                                  ↳ QDP

                                                                    ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

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 0 SCCs with 1 less node.