Termination w.r.t. Q proof of /tmp/tmpI4C2l6/otto03.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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fac(x) → help(x, 0)
help(x, c) → if(lt(c, x), x, c)
if(true, x, c) → times(s(c), help(x, s(c)))
if(false, x, c) → s(0)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fac(x) → help(x, 0)
help(x, c) → if(lt(c, x), x, c)
if(true, x, c) → times(s(c), help(x, s(c)))
if(false, x, c) → s(0)

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:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fac(x) → help(x, 0)
help(x, c) → if(lt(c, x), x, c)
if(true, x, c) → times(s(c), help(x, s(c)))
if(false, x, c) → s(0)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fac(x0)
help(x0, x1)
if(true, x0, x1)
if(false, x0, 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:

HELP(x, c) → IF(lt(c, x), x, c)
LT(s(x), s(y)) → LT(x, y)
FAC(x) → HELP(x, 0)
HELP(x, c) → LT(c, x)
IF(true, x, c) → HELP(x, s(c))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fac(x) → help(x, 0)
help(x, c) → if(lt(c, x), x, c)
if(true, x, c) → times(s(c), help(x, s(c)))
if(false, x, c) → s(0)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fac(x0)
help(x0, x1)
if(true, x0, x1)
if(false, x0, 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:

HELP(x, c) → IF(lt(c, x), x, c)
LT(s(x), s(y)) → LT(x, y)
FAC(x) → HELP(x, 0)
HELP(x, c) → LT(c, x)
IF(true, x, c) → HELP(x, s(c))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fac(x) → help(x, 0)
help(x, c) → if(lt(c, x), x, c)
if(true, x, c) → times(s(c), help(x, s(c)))
if(false, x, c) → s(0)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fac(x0)
help(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fac(x) → help(x, 0)
help(x, c) → if(lt(c, x), x, c)
if(true, x, c) → times(s(c), help(x, s(c)))
if(false, x, c) → s(0)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fac(x0)
help(x0, x1)
if(true, x0, x1)
if(false, x0, 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:

LT(s(x), s(y)) → LT(x, y)

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

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fac(x0)
help(x0, x1)
if(true, x0, x1)
if(false, x0, 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.

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fac(x0)
help(x0, x1)
if(true, x0, x1)
if(false, x0, 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:

LT(s(x), s(y)) → LT(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:

LT(s(x), s(y)) → LT(x, y)
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:

HELP(x, c) → IF(lt(c, x), x, c)
IF(true, x, c) → HELP(x, s(c))

The TRS R consists of the following rules:

lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
fac(x) → help(x, 0)
help(x, c) → if(lt(c, x), x, c)
if(true, x, c) → times(s(c), help(x, s(c)))
if(false, x, c) → s(0)

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fac(x0)
help(x0, x1)
if(true, x0, x1)
if(false, x0, 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:

HELP(x, c) → IF(lt(c, x), x, c)
IF(true, x, c) → HELP(x, s(c))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
fac(x0)
help(x0, x1)
if(true, x0, x1)
if(false, x0, 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.

fac(x0)
help(x0, x1)
if(true, x0, x1)
if(false, 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:

HELP(x, c) → IF(lt(c, x), x, c)
IF(true, x, c) → HELP(x, s(c))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(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 HELP(x, c) → IF(lt(c, x), x, c) the following chains were created:

We consider the chain IF(true, x, c) → HELP(x, s(c)), HELP(x, c) → IF(lt(c, x), x, c) which results in the following constraint:
(1) (HELP(x2, s(x3))=HELP(x4, x5) ⇒ HELP(x4, x5)≥IF(lt(x5, x4), x4, x5))

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

For Pair IF(true, x, c) → HELP(x, s(c)) the following chains were created:

We consider the chain HELP(x, c) → IF(lt(c, x), x, c), IF(true, x, c) → HELP(x, s(c)) which results in the following constraint:
(3)    (IF(lt(x7, x6), x6, x7)=IF(true, x8, x9) ⇒ IF(true, x8, x9)≥HELP(x8, s(x9)))

We simplified constraint (3) using rules (I), (II), (III) which results in the following new constraint:
(4)    (lt(x7, x6)=true ⇒ IF(true, x6, x7)≥HELP(x6, s(x7)))

We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on lt(x7, x6)=true which results in the following new constraints:
(5)    (true=true ⇒ IF(true, s(x12), 0)≥HELP(s(x12), s(0)))

(6)    (lt(x14, x15)=true∧(lt(x14, x15)=true ⇒ IF(true, x15, x14)≥HELP(x15, s(x14))) ⇒ IF(true, s(x15), s(x14))≥HELP(s(x15), s(s(x14))))

We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
(7)    (IF(true, s(x12), 0)≥HELP(s(x12), s(0)))

We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (lt(x14, x15)=true ⇒ IF(true, x15, x14)≥HELP(x15, s(x14))) with σ = [ ] which results in the following new constraint:
(8)    (IF(true, x15, x14)≥HELP(x15, s(x14)) ⇒ IF(true, s(x15), s(x14))≥HELP(s(x15), s(s(x14))))

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

HELP(x, c) → IF(lt(c, x), x, c)
- (HELP(x2, s(x3))≥IF(lt(s(x3), x2), x2, s(x3)))
IF(true, x, c) → HELP(x, s(c))
- (IF(true, s(x12), 0)≥HELP(s(x12), s(0)))
- (IF(true, x15, x14)≥HELP(x15, s(x14)) ⇒ IF(true, s(x15), s(x14))≥HELP(s(x15), s(s(x14))))

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(HELP(x₁, x₂)) = -1 + x₁ - x₂
POL(IF(x₁, x₂, x₃)) = -1 - x₁ + x₂ - x₃
POL(c) = -2
POL(false) = 1
POL(lt(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

IF(true, x, c) → HELP(x, s(c))

The following pairs are in P_bound:

IF(true, x, c) → HELP(x, s(c))

The following rules are usable:

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

↳ 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:

HELP(x, c) → IF(lt(c, x), x, c)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(0, s(x0))
lt(x0, 0)
lt(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.