Termination w.r.t. Q proof of /tmp/tmpBfRLKu/cade05t.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, 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)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

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)

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(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:

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(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) → MIN(x, y)
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) → EQUAL(min(x, y), y)
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

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:

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(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) → MIN(x, y)
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) → EQUAL(min(x, y), y)
MINUS(x, y) → COND(equal(min(x, y), y), x, y)

The TRS R consists of the following rules:

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:

minus(x0, x1)
cond(true, x0, x1)
min(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(s(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 2 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:

EQUAL(s(x), s(y)) → EQUAL(x, y)

The TRS R consists of the following rules:

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:

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

              ↳ 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:

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

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 [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:

EQUAL(s(x), s(y)) → EQUAL(x, y)
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:

MIN(s(u), s(v)) → MIN(u, v)

The TRS R consists of the following rules:

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:

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

MIN(s(u), s(v)) → MIN(u, v)

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

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

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

  ↳ Overlay + Local Confluence

    ↳ 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, 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:

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

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:

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

↳ 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) → 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(0, x0)
min(x0, 0)
min(s(x0), s(x1))
equal(0, 0)
equal(s(x0), 0)
equal(0, s(x0))
equal(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 COND(true, x, y) → MINUS(x, s(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:
(1)    (COND(equal(min(x2, x3), x3), x2, x3)=COND(true, x4, x5) ⇒ COND(true, x4, x5)≥MINUS(x4, s(x5)))

We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
(2)    (min(x2, x3)=x12∧equal(x12, x3)=true ⇒ COND(true, x2, x3)≥MINUS(x2, s(x3)))

We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on equal(x12, x3)=true which results in the following new constraints:
(3)    (true=true∧min(x2, 0)=0 ⇒ COND(true, x2, 0)≥MINUS(x2, s(0)))

(4)    (equal(x15, x16)=true∧min(x2, s(x16))=s(x15)∧(∀x17:equal(x15, x16)=true∧min(x17, x16)=x15 ⇒ COND(true, x17, x16)≥MINUS(x17, s(x16))) ⇒ COND(true, x2, s(x16))≥MINUS(x2, s(s(x16))))

We simplified constraint (3) using rules (I), (II), (VII) which results in the following new constraint:
(5)    (0=x18∧min(x2, x18)=0 ⇒ COND(true, x2, 0)≥MINUS(x2, s(0)))

We simplified constraint (4) using rule (VII) which results in the following new constraint:
(6)    (equal(x15, x16)=true∧s(x16)=x23∧min(x2, x23)=s(x15)∧(∀x17:equal(x15, x16)=true∧min(x17, x16)=x15 ⇒ COND(true, x17, x16)≥MINUS(x17, s(x16))) ⇒ COND(true, x2, s(x16))≥MINUS(x2, s(s(x16))))

We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on min(x2, x18)=0 which results in the following new constraints:
(7)    (0=0∧0=x19 ⇒ COND(true, 0, 0)≥MINUS(0, s(0)))

(8)    (0=0∧0=0 ⇒ COND(true, x20, 0)≥MINUS(x20, s(0)))

We simplified constraint (7) using rules (I), (II), (IV) which results in the following new constraint:
(9)    (COND(true, 0, 0)≥MINUS(0, s(0)))

We simplified constraint (8) using rules (I), (II) which results in the following new constraint:
(10)    (COND(true, x20, 0)≥MINUS(x20, s(0)))

We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on min(x2, x23)=s(x15) which results in the following new constraint:
(11)    (s(min(x26, x27))=s(x15)∧equal(x15, x16)=true∧s(x16)=s(x27)∧(∀x17:equal(x15, x16)=true∧min(x17, x16)=x15 ⇒ COND(true, x17, x16)≥MINUS(x17, s(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 ⇒ COND(true, x30, x29)≥MINUS(x30, s(x29))) ⇒ COND(true, x26, s(x29))≥MINUS(x26, s(s(x29)))) ⇒ COND(true, s(x26), s(x16))≥MINUS(s(x26), s(s(x16))))

We simplified constraint (11) using rules (I), (II), (III) which results in the following new constraint:
(12)    (min(x26, x27)=x15∧equal(x15, x27)=true∧(∀x17:equal(x15, x27)=true∧min(x17, x27)=x15 ⇒ COND(true, x17, x27)≥MINUS(x17, s(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 ⇒ COND(true, x30, x29)≥MINUS(x30, s(x29))) ⇒ COND(true, x26, s(x29))≥MINUS(x26, s(s(x29)))) ⇒ COND(true, s(x26), s(x27))≥MINUS(s(x26), s(s(x27))))

We simplified constraint (12) using rule (VI) where we applied the induction hypothesis (∀x17:equal(x15, x27)=true∧min(x17, x27)=x15 ⇒ COND(true, x17, x27)≥MINUS(x17, s(x27))) with σ = [x17 / x26] which results in the following new constraint:
(13)    (COND(true, x26, x27)≥MINUS(x26, s(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 ⇒ COND(true, x30, x29)≥MINUS(x30, s(x29))) ⇒ COND(true, x26, s(x29))≥MINUS(x26, s(s(x29)))) ⇒ COND(true, s(x26), s(x27))≥MINUS(s(x26), s(s(x27))))

We simplified constraint (13) using rule (IV) which results in the following new constraint:
(14)    (COND(true, x26, x27)≥MINUS(x26, s(x27)) ⇒ COND(true, s(x26), s(x27))≥MINUS(s(x26), s(s(x27))))

For Pair MINUS(x, y) → COND(equal(min(x, y), y), x, 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:
(15) (MINUS(x6, s(x7))=MINUS(x8, x9) ⇒ MINUS(x8, x9)≥COND(equal(min(x8, x9), x9), x8, x9))

We simplified constraint (15) using rules (I), (II), (III) which results in the following new constraint:
(16) (MINUS(x6, s(x7))≥COND(equal(min(x6, s(x7)), s(x7)), x6, s(x7)))

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

COND(true, x, y) → MINUS(x, s(y))
- (COND(true, 0, 0)≥MINUS(0, s(0)))
- (COND(true, x20, 0)≥MINUS(x20, s(0)))
- (COND(true, x26, x27)≥MINUS(x26, s(x27)) ⇒ COND(true, s(x26), s(x27))≥MINUS(s(x26), s(s(x27))))
MINUS(x, y) → COND(equal(min(x, y), y), x, y)
- (MINUS(x6, s(x7))≥COND(equal(min(x6, s(x7)), s(x7)), x6, s(x7)))

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(equal(x₁, x₂)) = 0
POL(false) = 0
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:

COND(true, x, y) → MINUS(x, s(y))

The following rules are usable:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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