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

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(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 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(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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

IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))
LT(s(x), s(y)) → LT(x, y)
IF(true, x, s(y), c) → PLUS(c, s(y))
PLUS(x, s(y)) → PLUS(x, y)
HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
HELP(x, s(y), c) → LT(c, x)
QUOT(x, s(y)) → HELP(x, s(y), 0)

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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:

IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))
LT(s(x), s(y)) → LT(x, y)
IF(true, x, s(y), c) → PLUS(c, s(y))
PLUS(x, s(y)) → PLUS(x, y)
HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
HELP(x, s(y), c) → LT(c, x)
QUOT(x, s(y)) → HELP(x, s(y), 0)

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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(x, s(y)) → PLUS(x, y)

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

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

↳ 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(x, s(y)) → PLUS(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:

PLUS(x, s(y)) → PLUS(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:

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

↳ 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:

LT(s(x), s(y)) → LT(x, y)

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

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

↳ 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:

LT(s(x), s(y)) → LT(x, y)

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

              ↳ QDP

                ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

↳ 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(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))
HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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.

quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y))) at position [2] we obtained the following new rules:

IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ NonInfProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We consider the chain IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y))), HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c) which results in the following constraint:
(1) (HELP(x3, s(x4), s(plus(x5, x4)))=HELP(x6, s(x7), x8) ⇒ HELP(x6, s(x7), x8)≥IF(lt(x8, x6), x6, s(x7), x8))

We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint:
(2) (HELP(x3, s(x4), x8)≥IF(lt(x8, x3), x3, s(x4), x8))

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

We consider the chain HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c), IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y))) which results in the following constraint:
(3)    (IF(lt(x11, x9), x9, s(x10), x11)=IF(true, x12, s(x13), x14) ⇒ IF(true, x12, s(x13), x14)≥HELP(x12, s(x13), s(plus(x14, x13))))

We simplified constraint (3) using rules (I), (II), (III) which results in the following new constraint:
(4)    (lt(x11, x9)=true ⇒ IF(true, x9, s(x10), x11)≥HELP(x9, s(x10), s(plus(x11, x10))))

We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on lt(x11, x9)=true which results in the following new constraints:
(5)    (true=true ⇒ IF(true, s(x19), s(x10), 0)≥HELP(s(x19), s(x10), s(plus(0, x10))))

(6)    (lt(x20, x21)=true∧(∀x22:lt(x20, x21)=true ⇒ IF(true, x21, s(x22), x20)≥HELP(x21, s(x22), s(plus(x20, x22)))) ⇒ IF(true, s(x21), s(x10), s(x20))≥HELP(s(x21), s(x10), s(plus(s(x20), x10))))

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

We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (∀x22:lt(x20, x21)=true ⇒ IF(true, x21, s(x22), x20)≥HELP(x21, s(x22), s(plus(x20, x22)))) with σ = [x22 / x10] which results in the following new constraint:
(8)    (IF(true, x21, s(x10), x20)≥HELP(x21, s(x10), s(plus(x20, x10))) ⇒ IF(true, s(x21), s(x10), s(x20))≥HELP(s(x21), s(x10), s(plus(s(x20), x10))))

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

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
- (HELP(x3, s(x4), x8)≥IF(lt(x8, x3), x3, s(x4), x8))
IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))
- (IF(true, s(x19), s(x10), 0)≥HELP(s(x19), s(x10), s(plus(0, x10))))
- (IF(true, x21, s(x10), x20)≥HELP(x21, s(x10), s(plus(x20, x10))) ⇒ IF(true, s(x21), s(x10), s(x20))≥HELP(s(x21), s(x10), s(plus(s(x20), x10))))

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, x₃)) = -1 + x₁ + x₂ - x₃
POL(IF(x₁, x₂, x₃, x₄)) = -1 - x₁ + x₂ + x₃ - x₄
POL(c) = -1
POL(false) = 0
POL(lt(x₁, x₂)) = 0
POL(plus(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))

The following pairs are in P_bound:

IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))

The following rules are usable:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ NonInfProof

                              ↳ QDP

                                ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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