Termination w.r.t. Q proof of /tmp/tmp0DIUaO/ex13.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:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))

The set Q consists of the following terms:

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

↳ QTRS

  ↳ DependencyPairsProof

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

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))

The set Q consists of the following terms:

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

PLUS(plus(n, m), u) → PLUS(n, plus(m, u))
COND(true, x, y, z) → PLUS(s(y), z)
PLUS(plus(n, m), u) → PLUS(m, u)
COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)
DIV(x, s(y)) → D(x, s(y), 0)
D(x, s(y), z) → GE(x, z)
PLUS(n, s(m)) → PLUS(n, m)
GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))

The set Q consists of the following terms:

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

PLUS(plus(n, m), u) → PLUS(n, plus(m, u))
COND(true, x, y, z) → PLUS(s(y), z)
PLUS(plus(n, m), u) → PLUS(m, u)
COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)
DIV(x, s(y)) → D(x, s(y), 0)
D(x, s(y), z) → GE(x, z)
PLUS(n, s(m)) → PLUS(n, m)
GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))

The set Q consists of the following terms:

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

PLUS(plus(n, m), u) → PLUS(n, plus(m, u))
PLUS(plus(n, m), u) → PLUS(m, u)
PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))

The set Q consists of the following terms:

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

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 [13] we can delete all non-usable rules [14] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

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

PLUS(plus(n, m), u) → PLUS(n, plus(m, u))
PLUS(plus(n, m), u) → PLUS(m, u)
PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))

The set Q consists of the following terms:

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

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.

ge(x0, 0)
d(x0, s(x1), x2)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

PLUS(plus(n, m), u) → PLUS(n, plus(m, u))
PLUS(plus(n, m), u) → PLUS(m, u)
PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))

The set Q consists of the following terms:

plus(x0, s(x1))
plus(x0, 0)
plus(plus(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [16] together with the size-change analysis [27] 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:

PLUS(n, s(m)) → PLUS(n, m)
The graph contains the following edges 1 >= 1, 2 > 2
PLUS(plus(n, m), u) → PLUS(m, u)
The graph contains the following edges 1 > 1, 2 >= 2
PLUS(plus(n, m), u) → PLUS(n, plus(m, u))
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))

The set Q consists of the following terms:

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

GE(s(u), s(v)) → GE(u, v)

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

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

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.

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

GE(s(u), s(v)) → GE(u, v)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [16] together with the size-change analysis [27] 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:

GE(s(u), s(v)) → GE(u, v)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))

The set Q consists of the following terms:

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
d(x0, s(x1), x2)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))
plus(plus(x0, x1), x2)

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.

d(x0, s(x1), x2)
cond(false, x0, x1, x2)
cond(true, x0, x1, x2)
div(x0, s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

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

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(plus(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule D(x, s(y), z) → COND(ge(x, z), x, y, z) at position [0] we obtained the following new rules:

D(x0, s(y1), 0) → COND(true, x0, y1, 0)
D(0, s(y1), s(x0)) → COND(false, 0, y1, s(x0))
D(s(x0), s(y1), s(x1)) → COND(ge(x0, x1), s(x0), y1, s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                    ↳ NonInfProof

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

D(s(x0), s(y1), s(x1)) → COND(ge(x0, x1), s(x0), y1, s(x1))
COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x0, s(y1), 0) → COND(true, x0, y1, 0)
D(0, s(y1), s(x0)) → COND(false, 0, y1, s(x0))

The TRS R consists of the following rules:

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(plus(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                    ↳ NonInfProof

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

D(s(x0), s(y1), s(x1)) → COND(ge(x0, x1), s(x0), y1, s(x1))
COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x0, s(y1), 0) → COND(true, x0, y1, 0)

The TRS R consists of the following rules:

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(plus(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule COND(true, x, y, z) → D(x, s(y), plus(s(y), z)) at position [2] we obtained the following new rules:

COND(true, y0, y1, s(x1)) → D(y0, s(y1), s(plus(s(y1), x1)))
COND(true, y0, y1, 0) → D(y0, s(y1), s(y1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                    ↳ NonInfProof

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

COND(true, y0, y1, s(x1)) → D(y0, s(y1), s(plus(s(y1), x1)))
COND(true, y0, y1, 0) → D(y0, s(y1), s(y1))
D(s(x0), s(y1), s(x1)) → COND(ge(x0, x1), s(x0), y1, s(x1))
D(x0, s(y1), 0) → COND(true, x0, y1, 0)

The TRS R consists of the following rules:

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(plus(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                    ↳ NonInfProof

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

COND(true, y0, y1, s(x1)) → D(y0, s(y1), s(plus(s(y1), x1)))
D(s(x0), s(y1), s(x1)) → COND(ge(x0, x1), s(x0), y1, s(x1))

The TRS R consists of the following rules:

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(plus(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND(true, y0, y1, s(x1)) → D(y0, s(y1), s(plus(s(y1), x1))) we obtained the following new rules:

COND(true, s(z0), z1, s(z2)) → D(s(z0), s(z1), s(plus(s(z1), z2)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                    ↳ NonInfProof

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

COND(true, s(z0), z1, s(z2)) → D(s(z0), s(z1), s(plus(s(z1), z2)))
D(s(x0), s(y1), s(x1)) → COND(ge(x0, x1), s(x0), y1, s(x1))

The TRS R consists of the following rules:

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(plus(x0, x1), x2)

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

For Pair COND(true, x, y, z) → D(x, s(y), plus(s(y), z)) the following chains were created:

We consider the chain COND(true, x, y, z) → D(x, s(y), plus(s(y), z)), D(x, s(y), z) → COND(ge(x, z), x, y, z) which results in the following constraint:
- (1) (D(x3, s(x4), plus(s(x4), x5))=D(x6, s(x7), x8) ⇒ COND(true, x3, x4, x5)≥D(x3, s(x4), plus(s(x4), x5)))
We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
- (2) (COND(true, x3, x4, x5)≥D(x3, s(x4), plus(s(x4), x5)))

For Pair D(x, s(y), z) → COND(ge(x, z), x, y, z) the following chains were created:

We consider the chain D(x, s(y), z) → COND(ge(x, z), x, y, z), COND(true, x, y, z) → D(x, s(y), plus(s(y), z)) which results in the following constraint:
- (3) (COND(ge(x9, x11), x9, x10, x11)=COND(true, x12, x13, x14) ⇒ D(x9, s(x10), x11)≥COND(ge(x9, x11), x9, x10, x11))
We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:
- (4) (ge(x9, x11)=true ⇒ D(x9, s(x10), x11)≥COND(ge(x9, x11), x9, x10, x11))
We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on ge(x9, x11)=true which results in the following new constraints:
- (5) (true=true ⇒ D(x18, s(x10), 0)≥COND(ge(x18, 0), x18, x10, 0))
- (6) (ge(x20, x21)=true∧(∀x22:ge(x20, x21)=true ⇒ D(x20, s(x22), x21)≥COND(ge(x20, x21), x20, x22, x21)) ⇒ D(s(x20), s(x10), s(x21))≥COND(ge(s(x20), s(x21)), s(x20), x10, s(x21)))
We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
- (7) (D(x18, s(x10), 0)≥COND(ge(x18, 0), x18, x10, 0))
We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (∀x22:ge(x20, x21)=true ⇒ D(x20, s(x22), x21)≥COND(ge(x20, x21), x20, x22, x21)) with σ = [x22 / x10] which results in the following new constraint:
- (8) (D(x20, s(x10), x21)≥COND(ge(x20, x21), x20, x10, x21) ⇒ D(s(x20), s(x10), s(x21))≥COND(ge(s(x20), s(x21)), s(x20), x10, s(x21)))

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

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
- (COND(true, x3, x4, x5)≥D(x3, s(x4), plus(s(x4), x5)))
D(x, s(y), z) → COND(ge(x, z), x, y, z)
- (D(x18, s(x10), 0)≥COND(ge(x18, 0), x18, x10, 0))
- (D(x20, s(x10), x21)≥COND(ge(x20, x21), x20, x10, x21) ⇒ D(s(x20), s(x10), s(x21))≥COND(ge(s(x20), s(x21)), s(x20), x10, s(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 [21]:

POL(0) = 0
POL(COND(x₁, x₂, x₃, x₄)) = -1 - x₁ + 2·x₂ - x₄
POL(D(x₁, x₂, x₃)) = -1 + 2·x₁ + x₂ - x₃
POL(c) = -1
POL(false) = 0
POL(ge(x₁, x₂)) = 0
POL(plus(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = 2 + x₁
POL(true) = 0

The following pairs are in P_>:

D(x, s(y), z) → COND(ge(x, z), x, y, z)

The following pairs are in P_bound:

D(x, s(y), z) → COND(ge(x, z), x, y, z)

The following rules are usable:

s(plus(n, m)) → plus(n, s(m))
true → ge(u, 0)
ge(u, v) → ge(s(u), s(v))
n → plus(n, 0)
plus(n, plus(m, u)) → plus(plus(n, m), u)
false → ge(0, s(v))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))

The TRS R consists of the following rules:

plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
plus(plus(n, m), u) → plus(n, plus(m, u))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
plus(x0, s(x1))
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(plus(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.