Termination w.r.t. Q proof of /tmp/tmpJGMHau/ex04.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(min(x, y), x, y)
cond(y, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

min(s(x0), s(x1))
cond(x0, x1, x0)
minus(x0, x1)
min(x0, 0)
min(0, 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:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

min(s(x0), s(x1))
cond(x0, x1, x0)
minus(x0, x1)
min(x0, 0)
min(0, 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:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(min(x, y), x, y)
cond(y, x, y) → s(minus(x, s(y)))
min(0, v) → 0
min(u, 0) → 0
min(s(u), s(v)) → s(min(u, v))

The set Q consists of the following terms:

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

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))
cond(x0, x1, x0)
minus(x0, x1)
min(x0, 0)
min(0, 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))
cond(x0, x1, x0)
minus(x0, x1)
min(x0, 0)
min(0, x0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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)

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:

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

            ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

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

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

The set Q consists of the following terms:

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

cond(x0, x1, x0)
minus(x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

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

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

The set Q consists of the following terms:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                    ↳ NonInfProof

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

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

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))

The set Q consists of the following terms:

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

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

                    ↳ NonInfProof

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

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

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))

The set Q consists of the following terms:

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

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

MINUS(0, s(z0)) → COND(0, 0, s(z0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ DependencyGraphProof

                    ↳ NonInfProof

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

MINUS(0, s(z0)) → COND(0, 0, s(z0))
COND(y, x, y) → MINUS(x, s(y))
MINUS(s(x0), s(x1)) → COND(s(min(x0, 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))

The set Q consists of the following terms:

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

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                    ↳ NonInfProof

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

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

The set Q consists of the following terms:

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

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

COND(s(y_0), s(z0), s(y_0)) → MINUS(s(z0), s(s(y_0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                    ↳ NonInfProof

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

MINUS(s(x0), s(x1)) → COND(s(min(x0, x1)), s(x0), s(x1))
COND(s(y_0), s(z0), s(y_0)) → MINUS(s(z0), s(s(y_0)))

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))

The set Q consists of the following terms:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                    ↳ NonInfProof

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

MINUS(s(z1), s(s(z0))) → COND(s(min(z1, s(z0))), s(z1), s(s(z0)))
COND(s(y_0), s(z0), s(y_0)) → MINUS(s(z0), s(s(y_0)))

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))

The set Q consists of the following terms:

min(s(x0), s(x1))
min(x0, 0)
min(0, 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(y, x, y) → MINUS(x, s(y)) the following chains were created:

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

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

We consider the chain MINUS(x, y) → COND(min(x, y), x, y), COND(y, x, y) → MINUS(x, s(y)) which results in the following constraint:
- (3) (COND(min(x6, x7), x6, x7)=COND(x8, x9, x8) ⇒ MINUS(x6, x7)≥COND(min(x6, x7), x6, x7))
We simplified constraint (3) using rules (I), (II), (III), (IV) which results in the following new constraint:
- (4) (min(x6, x7)=x7 ⇒ MINUS(x6, x7)≥COND(min(x6, x7), x6, x7))
We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on min(x6, x7)=x7 which results in the following new constraints:
- (5) (0=x12 ⇒ MINUS(0, x12)≥COND(min(0, x12), 0, x12))
- (6) (0=0 ⇒ MINUS(x13, 0)≥COND(min(x13, 0), x13, 0))
- (7) (s(min(x14, x15))=s(x15)∧(min(x14, x15)=x15 ⇒ MINUS(x14, x15)≥COND(min(x14, x15), x14, x15)) ⇒ MINUS(s(x14), s(x15))≥COND(min(s(x14), s(x15)), s(x14), s(x15)))
We simplified constraint (5) using rule (III) which results in the following new constraint:
- (8) (MINUS(0, 0)≥COND(min(0, 0), 0, 0))
We simplified constraint (6) using rules (I), (II) which results in the following new constraint:
- (9) (MINUS(x13, 0)≥COND(min(x13, 0), x13, 0))
We simplified constraint (7) using rules (I), (II) which results in the following new constraint:
- (10) (min(x14, x15)=x15∧(min(x14, x15)=x15 ⇒ MINUS(x14, x15)≥COND(min(x14, x15), x14, x15)) ⇒ MINUS(s(x14), s(x15))≥COND(min(s(x14), s(x15)), s(x14), s(x15)))
We simplified constraint (10) using rule (VI) where we applied the induction hypothesis (min(x14, x15)=x15 ⇒ MINUS(x14, x15)≥COND(min(x14, x15), x14, x15)) with σ = [ ] which results in the following new constraint:
- (11) (MINUS(x14, x15)≥COND(min(x14, x15), x14, x15) ⇒ MINUS(s(x14), s(x15))≥COND(min(s(x14), s(x15)), s(x14), s(x15)))

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

COND(y, x, y) → MINUS(x, s(y))
- (COND(x2, x3, x2)≥MINUS(x3, s(x2)))
MINUS(x, y) → COND(min(x, y), x, y)
- (MINUS(0, 0)≥COND(min(0, 0), 0, 0))
- (MINUS(x13, 0)≥COND(min(x13, 0), x13, 0))
- (MINUS(x14, x15)≥COND(min(x14, x15), x14, x15) ⇒ MINUS(s(x14), s(x15))≥COND(min(s(x14), s(x15)), s(x14), s(x15)))

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

The following pairs are in P_>:

COND(y, x, y) → MINUS(x, s(y))

The following pairs are in P_bound:

MINUS(x, y) → COND(min(x, y), x, y)

There are no usable rules

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(min(x, 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))

The set Q consists of the following terms:

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

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Narrowing

                    ↳ NonInfProof

                      ↳ AND

                        ↳ QDP

                        ↳ QDP

                          ↳ DependencyGraphProof

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

COND(y, 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))

The set Q consists of the following terms:

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