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

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(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:

INT(x, y) → LE(x, y)
INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(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:

INT(x, y) → LE(x, y)
INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(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 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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(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:

LE(s(x), s(y)) → LE(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(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.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(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:

LE(s(x), s(y)) → LE(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:

LE(s(x), s(y)) → LE(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:

INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(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:

INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(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.

int(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:

INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(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 INT(x, y) → IF(le(x, y), x, y) the following chains were created:

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

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

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

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

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

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

(6)    (le(x14, x15)=true∧(le(x14, x15)=true ⇒ IF(true, x14, x15)≥INT(s(x14), x15)) ⇒ IF(true, s(x14), s(x15))≥INT(s(s(x14)), s(x15)))

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

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

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

INT(x, y) → IF(le(x, y), x, y)
- (INT(s(x2), x3)≥IF(le(s(x2), x3), s(x2), x3))
IF(true, x, y) → INT(s(x), y)
- (IF(true, 0, x12)≥INT(s(0), x12))
- (IF(true, x14, x15)≥INT(s(x14), x15) ⇒ IF(true, s(x14), s(x15))≥INT(s(s(x14)), s(x15)))

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(IF(x₁, x₂, x₃)) = -x₁ - x₂ + x₃
POL(INT(x₁, x₂)) = 1 - x₁ + x₂
POL(c) = -1
POL(false) = 0
POL(le(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

INT(x, y) → IF(le(x, y), x, y)

The following pairs are in P_bound:

IF(true, x, y) → INT(s(x), y)

The following rules are usable:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ AND

                            ↳ QDP

                              ↳ DependencyGraphProof

                            ↳ QDP

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

IF(true, x, y) → INT(s(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(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.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ AND

                            ↳ QDP

                            ↳ QDP

                              ↳ DependencyGraphProof

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

INT(x, y) → IF(le(x, y), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(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.