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

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))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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))

The set Q consists of the following terms:

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

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The set Q consists of the following terms:

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

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))

The set Q consists of the following terms:

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

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

PLUS(n, s(m)) → PLUS(n, m)

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

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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(n, s(m)) → PLUS(n, m)

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:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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:

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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)

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

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))

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))

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

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))

The TRS R consists of the following rules:

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

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ 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))
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)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(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 D(x, s(y), z) → COND(ge(x, z), x, 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) ⇒ D(x6, s(x7), x8)≥COND(ge(x6, x8), x6, x7, x8))

We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint:
(2) (D(x3, s(x4), x8)≥COND(ge(x3, x8), x3, x4, x8))

For Pair COND(true, x, y, z) → D(x, s(y), plus(s(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) ⇒ COND(true, x12, x13, x14)≥D(x12, s(x13), plus(s(x13), x14)))

We simplified constraint (3) using rules (I), (II), (III) which results in the following new constraint:
(4)    (ge(x9, x11)=true ⇒ COND(true, x9, x10, x11)≥D(x9, s(x10), plus(s(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 ⇒ COND(true, x19, x10, 0)≥D(x19, s(x10), plus(s(x10), 0)))

(6)    (ge(x21, x22)=true∧(∀x23:ge(x21, x22)=true ⇒ COND(true, x21, x23, x22)≥D(x21, s(x23), plus(s(x23), x22))) ⇒ COND(true, s(x21), x10, s(x22))≥D(s(x21), s(x10), plus(s(x10), s(x22))))

We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
(7)    (COND(true, x19, x10, 0)≥D(x19, s(x10), plus(s(x10), 0)))

We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (∀x23:ge(x21, x22)=true ⇒ COND(true, x21, x23, x22)≥D(x21, s(x23), plus(s(x23), x22))) with σ = [x23 / x10] which results in the following new constraint:
(8)    (COND(true, x21, x10, x22)≥D(x21, s(x10), plus(s(x10), x22)) ⇒ COND(true, s(x21), x10, s(x22))≥D(s(x21), s(x10), plus(s(x10), s(x22))))

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

D(x, s(y), z) → COND(ge(x, z), x, y, z)
- (D(x3, s(x4), x8)≥COND(ge(x3, x8), x3, x4, x8))
COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
- (COND(true, x19, x10, 0)≥D(x19, s(x10), plus(s(x10), 0)))
- (COND(true, x21, x10, x22)≥D(x21, s(x10), plus(s(x10), x22)) ⇒ COND(true, s(x21), x10, s(x22))≥D(s(x21), s(x10), plus(s(x10), s(x22))))

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

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

The following rules are usable:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ AND

                            ↳ QDP

                              ↳ DependencyGraphProof

                            ↳ QDP

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ AND

                            ↳ QDP

                            ↳ QDP

                              ↳ DependencyGraphProof

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

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