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

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

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

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

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:

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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:

minus(x0, x1)
cond(false, x0, x1)
cond(true, x0, x1)
ge(x0, 0)
ge(0, s(x0))
ge(s(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:

MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))
GE(s(u), s(v)) → GE(u, v)
MINUS(x, y) → GE(x, s(y))

The TRS R consists of the following rules:

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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:

minus(x0, x1)
cond(false, x0, x1)
cond(true, x0, x1)
ge(x0, 0)
ge(0, s(x0))
ge(s(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:

MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))
GE(s(u), s(v)) → GE(u, v)
MINUS(x, y) → GE(x, s(y))

The TRS R consists of the following rules:

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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:

minus(x0, x1)
cond(false, x0, x1)
cond(true, x0, x1)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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:

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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:

minus(x0, x1)
cond(false, x0, x1)
cond(true, x0, x1)
ge(x0, 0)
ge(0, s(x0))
ge(s(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

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:

minus(x0, x1)
cond(false, x0, x1)
cond(true, x0, x1)
ge(x0, 0)
ge(0, s(x0))
ge(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.

minus(x0, x1)
cond(false, x0, x1)
cond(true, x0, x1)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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 [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:

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

                ↳ UsableRulesProof

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

MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
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:

minus(x0, x1)
cond(false, x0, x1)
cond(true, x0, x1)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true

The set Q consists of the following terms:

minus(x0, x1)
cond(false, x0, x1)
cond(true, x0, x1)
ge(x0, 0)
ge(0, s(x0))
ge(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.

minus(x0, x1)
cond(false, x0, x1)
cond(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

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

MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true

The set Q consists of the following terms:

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

We consider the chain COND(true, x, y) → MINUS(x, s(y)), MINUS(x, y) → COND(ge(x, s(y)), x, y) which results in the following constraint:
(1) (MINUS(x2, s(x3))=MINUS(x4, x5) ⇒ MINUS(x4, x5)≥COND(ge(x4, s(x5)), x4, x5))

We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2) (MINUS(x2, s(x3))≥COND(ge(x2, s(s(x3))), x2, s(x3)))

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

We consider the chain MINUS(x, y) → COND(ge(x, s(y)), x, y), COND(true, x, y) → MINUS(x, s(y)) which results in the following constraint:
(3)    (COND(ge(x6, s(x7)), x6, x7)=COND(true, x8, x9) ⇒ COND(true, x8, x9)≥MINUS(x8, s(x9)))

We simplified constraint (3) using rules (I), (II), (III), (VII) which results in the following new constraint:
(4)    (s(x7)=x12∧ge(x6, x12)=true ⇒ COND(true, x6, x7)≥MINUS(x6, s(x7)))

We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on ge(x6, x12)=true which results in the following new constraints:
(5)    (ge(x14, x15)=true∧s(x7)=s(x15)∧(∀x16:ge(x14, x15)=true∧s(x16)=x15 ⇒ COND(true, x14, x16)≥MINUS(x14, s(x16))) ⇒ COND(true, s(x14), x7)≥MINUS(s(x14), s(x7)))

(6)    (true=true∧s(x7)=0 ⇒ COND(true, x17, x7)≥MINUS(x17, s(x7)))

We simplified constraint (5) using rules (I), (II), (III), (IV) which results in the following new constraint:
(7)    (ge(x14, x15)=true ⇒ COND(true, s(x14), x15)≥MINUS(s(x14), s(x15)))

We solved constraint (6) using rules (I), (II).We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on ge(x14, x15)=true which results in the following new constraints:
(8)    (ge(x19, x20)=true∧(ge(x19, x20)=true ⇒ COND(true, s(x19), x20)≥MINUS(s(x19), s(x20))) ⇒ COND(true, s(s(x19)), s(x20))≥MINUS(s(s(x19)), s(s(x20))))

(9)    (true=true ⇒ COND(true, s(x21), 0)≥MINUS(s(x21), s(0)))

We simplified constraint (8) using rule (VI) where we applied the induction hypothesis (ge(x19, x20)=true ⇒ COND(true, s(x19), x20)≥MINUS(s(x19), s(x20))) with σ = [ ] which results in the following new constraint:
(10)    (COND(true, s(x19), x20)≥MINUS(s(x19), s(x20)) ⇒ COND(true, s(s(x19)), s(x20))≥MINUS(s(s(x19)), s(s(x20))))

We simplified constraint (9) using rules (I), (II) which results in the following new constraint:
(11)    (COND(true, s(x21), 0)≥MINUS(s(x21), s(0)))

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

MINUS(x, y) → COND(ge(x, s(y)), x, y)
- (MINUS(x2, s(x3))≥COND(ge(x2, s(s(x3))), x2, s(x3)))
COND(true, x, y) → MINUS(x, s(y))
- (COND(true, s(x19), x20)≥MINUS(s(x19), s(x20)) ⇒ COND(true, s(s(x19)), s(x20))≥MINUS(s(s(x19)), s(s(x20))))
- (COND(true, s(x21), 0)≥MINUS(s(x21), s(0)))

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) = 1
POL(COND(x₁, x₂, x₃)) = -1 - x₁ + x₂ - x₃
POL(MINUS(x₁, x₂)) = -1 + x₁ - x₂
POL(c) = -2
POL(false) = 1
POL(ge(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:

COND(true, x, y) → MINUS(x, s(y))

The following rules are usable:

ge(u, v) → ge(s(u), s(v))
false → ge(0, s(v))
true → ge(u, 0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

MINUS(x, y) → COND(ge(x, s(y)), x, y)

The TRS R consists of the following rules:

ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true

The set Q consists of the following terms:

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