Termination w.r.t. Q proof of /tmp/tmpXXcmis/thiemann41.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:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(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:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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:

TIMES(x, y) → GENERATE(x, y)
TIMES(x, y) → SUM(generate(x, y))
SUM(cons(s(x), xs)) → SUM(cons(x, xs))
GENERATE(x, y) → GEN(x, y, 0)
SUM(cons(0, xs)) → SUM(xs)
IF(false, x, y, z) → GEN(x, y, s(z))
GE(s(x), s(y)) → GE(x, y)
GEN(x, y, z) → GE(z, x)
GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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:

TIMES(x, y) → GENERATE(x, y)
TIMES(x, y) → SUM(generate(x, y))
SUM(cons(s(x), xs)) → SUM(cons(x, xs))
GENERATE(x, y) → GEN(x, y, 0)
SUM(cons(0, xs)) → SUM(xs)
IF(false, x, y, z) → GEN(x, y, s(z))
GE(s(x), s(y)) → GE(x, y)
GEN(x, y, z) → GE(z, x)
GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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 3 SCCs with 4 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:

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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

              ↳ QDP

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

GE(s(x), s(y)) → GE(x, y)

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

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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.

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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

              ↳ QDP

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

GE(s(x), s(y)) → GE(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:

GE(s(x), s(y)) → GE(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:

SUM(cons(s(x), xs)) → SUM(cons(x, xs))
SUM(cons(0, xs)) → SUM(xs)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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

              ↳ QDP

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

SUM(cons(s(x), xs)) → SUM(cons(x, xs))
SUM(cons(0, xs)) → SUM(xs)

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

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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.

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

              ↳ QDP

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

SUM(cons(s(x), xs)) → SUM(cons(x, xs))
SUM(cons(0, xs)) → SUM(xs)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

SUM(cons(s(x), xs)) → SUM(cons(x, xs))
SUM(cons(0, xs)) → SUM(xs)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(SUM(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s(x₁)) = 2·x₁

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ PisEmptyProof

              ↳ QDP

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

IF(false, x, y, z) → GEN(x, y, s(z))
GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

times(x, y) → sum(generate(x, y))
generate(x, y) → gen(x, y, 0)
gen(x, y, z) → if(ge(z, x), x, y, z)
if(true, x, y, z) → nil
if(false, x, y, z) → cons(y, gen(x, y, s(z)))
sum(nil) → 0
sum(cons(0, xs)) → sum(xs)
sum(cons(s(x), xs)) → s(sum(cons(x, xs)))
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

IF(false, x, y, z) → GEN(x, y, s(z))
GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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.

times(x0, x1)
generate(x0, x1)
gen(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
sum(nil)
sum(cons(0, x0))
sum(cons(s(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:

IF(false, x, y, z) → GEN(x, y, s(z))
GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

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

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 IF(false, x, y, z) → GEN(x, y, s(z)) the following chains were created:

We consider the chain GEN(x, y, z) → IF(ge(z, x), x, y, z), IF(false, x, y, z) → GEN(x, y, s(z)) which results in the following constraint:
(1)    (IF(ge(x5, x3), x3, x4, x5)=IF(false, x6, x7, x8) ⇒ IF(false, x6, x7, x8)≥GEN(x6, x7, s(x8)))

We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2)    (ge(x5, x3)=false ⇒ IF(false, x3, x4, x5)≥GEN(x3, x4, s(x5)))

We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x5, x3)=false which results in the following new constraints:
(3)    (false=false ⇒ IF(false, s(x19), x4, 0)≥GEN(s(x19), x4, s(0)))

(4)    (ge(x20, x21)=false∧(∀x22:ge(x20, x21)=false ⇒ IF(false, x21, x22, x20)≥GEN(x21, x22, s(x20))) ⇒ IF(false, s(x21), x4, s(x20))≥GEN(s(x21), x4, s(s(x20))))

We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
(5)    (IF(false, s(x19), x4, 0)≥GEN(s(x19), x4, s(0)))

We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x22:ge(x20, x21)=false ⇒ IF(false, x21, x22, x20)≥GEN(x21, x22, s(x20))) with σ = [x22 / x4] which results in the following new constraint:
(6)    (IF(false, x21, x4, x20)≥GEN(x21, x4, s(x20)) ⇒ IF(false, s(x21), x4, s(x20))≥GEN(s(x21), x4, s(s(x20))))

For Pair GEN(x, y, z) → IF(ge(z, x), x, y, z) the following chains were created:

We consider the chain IF(false, x, y, z) → GEN(x, y, s(z)), GEN(x, y, z) → IF(ge(z, x), x, y, z) which results in the following constraint:
(7) (GEN(x9, x10, s(x11))=GEN(x12, x13, x14) ⇒ GEN(x12, x13, x14)≥IF(ge(x14, x12), x12, x13, x14))

We simplified constraint (7) using rules (I), (II), (III) which results in the following new constraint:
(8) (GEN(x9, x10, s(x11))≥IF(ge(s(x11), x9), x9, x10, s(x11)))

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

IF(false, x, y, z) → GEN(x, y, s(z))
- (IF(false, s(x19), x4, 0)≥GEN(s(x19), x4, s(0)))
- (IF(false, x21, x4, x20)≥GEN(x21, x4, s(x20)) ⇒ IF(false, s(x21), x4, s(x20))≥GEN(s(x21), x4, s(s(x20))))
GEN(x, y, z) → IF(ge(z, x), x, y, z)
- (GEN(x9, x10, s(x11))≥IF(ge(s(x11), x9), x9, x10, s(x11)))

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(GEN(x₁, x₂, x₃)) = -1 + x₁ + x₂ - x₃
POL(IF(x₁, x₂, x₃, x₄)) = -x₁ + x₂ + x₃ - x₄
POL(c) = -2
POL(false) = 1
POL(ge(x₁, x₂)) = 1
POL(s(x₁)) = 1 + x₁
POL(true) = 1

The following pairs are in P_>:

IF(false, x, y, z) → GEN(x, y, s(z))

The following pairs are in P_bound:

IF(false, x, y, z) → GEN(x, y, s(z))

The following rules are usable:

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

↳ 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:

GEN(x, y, z) → IF(ge(z, x), x, y, z)

The TRS R consists of the following rules:

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

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.