Termination w.r.t. Q proof of /tmp/tmpPFBtUT/ex05.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:

minus(x, x) → 0
minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, s(x0))

↳ QTRS

  ↳ DependencyPairsProof

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

minus(x, x) → 0
minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, s(x0))

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

EQUAL(s(x), s(y)) → EQUAL(x, y)
MIN(s(u), s(v)) → MIN(u, v)
COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)
MINUS(x, y) → EQUAL(min(x, y), y)
MINUS(x, y) → MIN(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, s(x0))

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:

EQUAL(s(x), s(y)) → EQUAL(x, y)
MIN(s(u), s(v)) → MIN(u, v)
COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)
MINUS(x, y) → EQUAL(min(x, y), y)
MINUS(x, y) → MIN(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

EQUAL(s(x), s(y)) → EQUAL(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, 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 [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:

EQUAL(s(x), s(y)) → EQUAL(x, y)

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

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, 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.

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

EQUAL(s(x), s(y)) → EQUAL(x, y)

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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

MIN(s(u), s(v)) → MIN(u, v)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

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

MIN(s(u), s(v)) → MIN(u, v)

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

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, 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.

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

MIN(s(u), s(v)) → MIN(u, v)

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

MIN(s(u), s(v)) → MIN(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) → MINUS(x, s(y))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(x, y) → cond(equal(min(x, y), y), x, y)
cond(true, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, s(x0))

↳ 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) → MINUS(x, s(y))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
minus(x0, x1)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
cond(true, x0, x1)
equal(0, 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.

minus(x0, x1)
cond(true, x0, 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) → MINUS(x, s(y))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
equal(0, s(x0))

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

MINUS(0, x0) → COND(equal(0, x0), 0, x0)
MINUS(x0, 0) → COND(equal(0, 0), x0, 0)
MINUS(s(x0), s(x1)) → COND(equal(s(min(x0, x1)), s(x1)), s(x0), 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:

MINUS(x0, 0) → COND(equal(0, 0), x0, 0)
COND(true, x, y) → MINUS(x, s(y))
MINUS(0, x0) → COND(equal(0, x0), 0, x0)
MINUS(s(x0), s(x1)) → COND(equal(s(min(x0, x1)), s(x1)), s(x0), s(x1))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
equal(0, s(x0))

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:

COND(true, x, y) → MINUS(x, s(y))
MINUS(0, x0) → COND(equal(0, x0), 0, x0)
MINUS(s(x0), s(x1)) → COND(equal(s(min(x0, x1)), s(x1)), s(x0), s(x1))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
equal(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule MINUS(0, x0) → COND(equal(0, x0), 0, x0) at position [0] we obtained the following new rules:

MINUS(0, s(x0)) → COND(false, 0, s(x0))
MINUS(0, 0) → COND(true, 0, 0)

↳ 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, x, y) → MINUS(x, s(y))
MINUS(0, s(x0)) → COND(false, 0, s(x0))
MINUS(0, 0) → COND(true, 0, 0)
MINUS(s(x0), s(x1)) → COND(equal(s(min(x0, x1)), s(x1)), s(x0), s(x1))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
equal(0, s(x0))

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, x, y) → MINUS(x, s(y))
MINUS(s(x0), s(x1)) → COND(equal(s(min(x0, x1)), s(x1)), s(x0), s(x1))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
equal(0, s(x0))

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

COND(true, s(z0), s(z1)) → MINUS(s(z0), s(s(z1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                    ↳ NonInfProof

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

COND(true, s(z0), s(z1)) → MINUS(s(z0), s(s(z1)))
MINUS(s(x0), s(x1)) → COND(equal(s(min(x0, x1)), s(x1)), s(x0), s(x1))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
equal(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule MINUS(s(x0), s(x1)) → COND(equal(s(min(x0, x1)), s(x1)), s(x0), s(x1)) we obtained the following new rules:

MINUS(s(z0), s(s(z1))) → COND(equal(s(min(z0, s(z1))), s(s(z1))), s(z0), s(s(z1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                    ↳ NonInfProof

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

COND(true, s(z0), s(z1)) → MINUS(s(z0), s(s(z1)))
MINUS(s(z0), s(s(z1))) → COND(equal(s(min(z0, s(z1))), s(s(z1))), s(z0), s(s(z1)))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
equal(0, s(x0))

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) → MINUS(x, s(y)) the following chains were created:

We consider the chain COND(true, x, y) → MINUS(x, s(y)), MINUS(x, y) → COND(equal(min(x, y), y), x, y) which results in the following constraint:
- (1) (MINUS(x2, s(x3))=MINUS(x4, x5) ⇒ COND(true, x2, x3)≥MINUS(x2, s(x3)))
We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
- (2) (COND(true, x2, x3)≥MINUS(x2, s(x3)))

For Pair MINUS(x, y) → COND(equal(min(x, y), y), x, y) the following chains were created:

We consider the chain MINUS(x, y) → COND(equal(min(x, y), y), x, y), COND(true, x, y) → MINUS(x, s(y)) which results in the following constraint:
- (3) (COND(equal(min(x6, x7), x7), x6, x7)=COND(true, x8, x9) ⇒ MINUS(x6, x7)≥COND(equal(min(x6, x7), x7), x6, x7))
We simplified constraint (3) using rules (I), (II), (IV), (VII) which results in the following new constraint:
- (4) (min(x6, x7)=x12∧equal(x12, x7)=true ⇒ MINUS(x6, x7)≥COND(equal(min(x6, x7), x7), x6, x7))
We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on equal(x12, x7)=true which results in the following new constraints:
- (5) (true=true∧min(x6, 0)=0 ⇒ MINUS(x6, 0)≥COND(equal(min(x6, 0), 0), x6, 0))
- (6) (equal(x15, x16)=true∧min(x6, s(x16))=s(x15)∧(∀x17:equal(x15, x16)=true∧min(x17, x16)=x15 ⇒ MINUS(x17, x16)≥COND(equal(min(x17, x16), x16), x17, x16)) ⇒ MINUS(x6, s(x16))≥COND(equal(min(x6, s(x16)), s(x16)), x6, s(x16)))
We simplified constraint (5) using rules (I), (II), (VII) which results in the following new constraint:
- (7) (0=x18∧min(x6, x18)=0 ⇒ MINUS(x6, 0)≥COND(equal(min(x6, 0), 0), x6, 0))
We simplified constraint (6) using rule (VII) which results in the following new constraint:
- (8) (equal(x15, x16)=true∧s(x16)=x23∧min(x6, x23)=s(x15)∧(∀x17:equal(x15, x16)=true∧min(x17, x16)=x15 ⇒ MINUS(x17, x16)≥COND(equal(min(x17, x16), x16), x17, x16)) ⇒ MINUS(x6, s(x16))≥COND(equal(min(x6, s(x16)), s(x16)), x6, s(x16)))
We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on min(x6, x18)=0 which results in the following new constraints:
- (9) (0=0∧0=x19 ⇒ MINUS(0, 0)≥COND(equal(min(0, 0), 0), 0, 0))
- (10) (0=0∧0=0 ⇒ MINUS(x20, 0)≥COND(equal(min(x20, 0), 0), x20, 0))
We simplified constraint (9) using rules (I), (II), (IV) which results in the following new constraint:
- (11) (MINUS(0, 0)≥COND(equal(min(0, 0), 0), 0, 0))
We simplified constraint (10) using rules (I), (II) which results in the following new constraint:
- (12) (MINUS(x20, 0)≥COND(equal(min(x20, 0), 0), x20, 0))
We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on min(x6, x23)=s(x15) which results in the following new constraint:
- (13) (s(min(x26, x27))=s(x15)∧equal(x15, x16)=true∧s(x16)=s(x27)∧(∀x17:equal(x15, x16)=true∧min(x17, x16)=x15 ⇒ MINUS(x17, x16)≥COND(equal(min(x17, x16), x16), x17, x16))∧(∀x28,x29,x30:min(x26, x27)=s(x28)∧equal(x28, x29)=true∧s(x29)=x27∧(∀x30:equal(x28, x29)=true∧min(x30, x29)=x28 ⇒ MINUS(x30, x29)≥COND(equal(min(x30, x29), x29), x30, x29)) ⇒ MINUS(x26, s(x29))≥COND(equal(min(x26, s(x29)), s(x29)), x26, s(x29))) ⇒ MINUS(s(x26), s(x16))≥COND(equal(min(s(x26), s(x16)), s(x16)), s(x26), s(x16)))
We simplified constraint (13) using rules (I), (II), (III) which results in the following new constraint:
- (14) (min(x26, x27)=x15∧equal(x15, x27)=true∧(∀x17:equal(x15, x27)=true∧min(x17, x27)=x15 ⇒ MINUS(x17, x27)≥COND(equal(min(x17, x27), x27), x17, x27))∧(∀x28,x29,x30:min(x26, x27)=s(x28)∧equal(x28, x29)=true∧s(x29)=x27∧(∀x30:equal(x28, x29)=true∧min(x30, x29)=x28 ⇒ MINUS(x30, x29)≥COND(equal(min(x30, x29), x29), x30, x29)) ⇒ MINUS(x26, s(x29))≥COND(equal(min(x26, s(x29)), s(x29)), x26, s(x29))) ⇒ MINUS(s(x26), s(x27))≥COND(equal(min(s(x26), s(x27)), s(x27)), s(x26), s(x27)))
We simplified constraint (14) using rule (VI) where we applied the induction hypothesis (∀x17:equal(x15, x27)=true∧min(x17, x27)=x15 ⇒ MINUS(x17, x27)≥COND(equal(min(x17, x27), x27), x17, x27)) with σ = [x17 / x26] which results in the following new constraint:
- (15) (MINUS(x26, x27)≥COND(equal(min(x26, x27), x27), x26, x27)∧(∀x28,x29,x30:min(x26, x27)=s(x28)∧equal(x28, x29)=true∧s(x29)=x27∧(∀x30:equal(x28, x29)=true∧min(x30, x29)=x28 ⇒ MINUS(x30, x29)≥COND(equal(min(x30, x29), x29), x30, x29)) ⇒ MINUS(x26, s(x29))≥COND(equal(min(x26, s(x29)), s(x29)), x26, s(x29))) ⇒ MINUS(s(x26), s(x27))≥COND(equal(min(s(x26), s(x27)), s(x27)), s(x26), s(x27)))
We simplified constraint (15) using rule (IV) which results in the following new constraint:
- (16) (MINUS(x26, x27)≥COND(equal(min(x26, x27), x27), x26, x27) ⇒ MINUS(s(x26), s(x27))≥COND(equal(min(s(x26), s(x27)), s(x27)), s(x26), s(x27)))

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

COND(true, x, y) → MINUS(x, s(y))
- (COND(true, x2, x3)≥MINUS(x2, s(x3)))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)
- (MINUS(0, 0)≥COND(equal(min(0, 0), 0), 0, 0))
- (MINUS(x20, 0)≥COND(equal(min(x20, 0), 0), x20, 0))
- (MINUS(x26, x27)≥COND(equal(min(x26, x27), x27), x26, x27) ⇒ MINUS(s(x26), s(x27))≥COND(equal(min(s(x26), s(x27)), s(x27)), s(x26), s(x27)))

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)) = -1 - x₁ + x₂ - x₃
POL(MINUS(x₁, x₂)) = -1 + x₁ - x₂
POL(c) = -1
POL(equal(x₁, x₂)) = 0
POL(false) = 2
POL(min(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

COND(true, x, y) → MINUS(x, s(y))

The following pairs are in P_bound:

MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The following rules are usable:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                          ↳ DependencyGraphProof

                        ↳ QDP

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

MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
equal(0, s(x0))

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

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

COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))
equal(0, 0) → true
equal(s(x), 0) → false
equal(0, s(y)) → false
equal(s(x), s(y)) → equal(x, y)

The set Q consists of the following terms:

min(s(x0), s(x1))
equal(s(x0), 0)
equal(0, 0)
min(x0, 0)
equal(s(x0), s(x1))
min(0, x0)
equal(0, s(x0))

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.