Termination w.r.t. Q proof of /tmp/tmpJIXJlc/thiemann22.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:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

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(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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(false, x, y, z) → LOOP(x, double(y), s(z))
DOUBLE(s(x)) → DOUBLE(x)
IF(false, x, y, z) → DOUBLE(y)
LOOP(x, s(y), z) → LE(x, s(y))
LOG(s(x)) → LOOP(s(x), s(0), 0)
LE(s(x), s(y)) → LE(x, y)
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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(false, x, y, z) → LOOP(x, double(y), s(z))
DOUBLE(s(x)) → DOUBLE(x)
IF(false, x, y, z) → DOUBLE(y)
LOOP(x, s(y), z) → LE(x, s(y))
LOG(s(x)) → LOOP(s(x), s(0), 0)
LE(s(x), s(y)) → LE(x, y)
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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:

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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:

DOUBLE(s(x)) → DOUBLE(x)

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

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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.

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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:

DOUBLE(s(x)) → DOUBLE(x)

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:

DOUBLE(s(x)) → DOUBLE(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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:

LE(s(x), s(y)) → LE(x, y)

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

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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.

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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:

LE(s(x), s(y)) → LE(x, y)

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

              ↳ QDP

                ↳ UsableRulesProof

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

IF(false, x, y, z) → LOOP(x, double(y), s(z))
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)

The TRS R consists of the following rules:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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(false, x, y, z) → LOOP(x, double(y), s(z))
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, 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.

log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)

↳ 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) → LOOP(x, double(y), s(z))
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))

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) → LOOP(x, double(y), s(z)) the following chains were created:

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

We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
(2)    (double(x4)=s(x7) ⇒ IF(false, x3, x4, x5)≥LOOP(x3, double(x4), s(x5)))

We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on double(x4)=s(x7) which results in the following new constraint:
(3)    (s(s(double(x18)))=s(x7)∧(∀x19,x20,x21:double(x18)=s(x19) ⇒ IF(false, x20, x18, x21)≥LOOP(x20, double(x18), s(x21))) ⇒ IF(false, x3, s(x18), x5)≥LOOP(x3, double(s(x18)), s(x5)))

We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:
(4)    (IF(false, x3, s(x18), x5)≥LOOP(x3, double(s(x18)), s(x5)))

For Pair LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z) the following chains were created:

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

We simplified constraint (5) using rules (I), (II), (IV), (VII) which results in the following new constraint:
(6)    (s(x10)=x22∧le(x9, x22)=false ⇒ LOOP(x9, s(x10), x11)≥IF(le(x9, s(x10)), x9, s(x10), x11))

We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on le(x9, x22)=false which results in the following new constraints:
(7)    (le(x24, x25)=false∧s(x10)=s(x25)∧(∀x26,x27:le(x24, x25)=false∧s(x26)=x25 ⇒ LOOP(x24, s(x26), x27)≥IF(le(x24, s(x26)), x24, s(x26), x27)) ⇒ LOOP(s(x24), s(x10), x11)≥IF(le(s(x24), s(x10)), s(x24), s(x10), x11))

(8)    (false=false∧s(x10)=0 ⇒ LOOP(s(x28), s(x10), x11)≥IF(le(s(x28), s(x10)), s(x28), s(x10), x11))

We simplified constraint (7) using rules (I), (II), (III), (IV) which results in the following new constraint:
(9)    (le(x24, x25)=false ⇒ LOOP(s(x24), s(x25), x11)≥IF(le(s(x24), s(x25)), s(x24), s(x25), x11))

We solved constraint (8) using rules (I), (II).We simplified constraint (9) using rule (V) (with possible (I) afterwards) using induction on le(x24, x25)=false which results in the following new constraints:
(10)    (le(x30, x31)=false∧(∀x32:le(x30, x31)=false ⇒ LOOP(s(x30), s(x31), x32)≥IF(le(s(x30), s(x31)), s(x30), s(x31), x32)) ⇒ LOOP(s(s(x30)), s(s(x31)), x11)≥IF(le(s(s(x30)), s(s(x31))), s(s(x30)), s(s(x31)), x11))

(11)    (false=false ⇒ LOOP(s(s(x33)), s(0), x11)≥IF(le(s(s(x33)), s(0)), s(s(x33)), s(0), x11))

We simplified constraint (10) using rule (VI) where we applied the induction hypothesis (∀x32:le(x30, x31)=false ⇒ LOOP(s(x30), s(x31), x32)≥IF(le(s(x30), s(x31)), s(x30), s(x31), x32)) with σ = [x32 / x11] which results in the following new constraint:
(12)    (LOOP(s(x30), s(x31), x11)≥IF(le(s(x30), s(x31)), s(x30), s(x31), x11) ⇒ LOOP(s(s(x30)), s(s(x31)), x11)≥IF(le(s(s(x30)), s(s(x31))), s(s(x30)), s(s(x31)), x11))

We simplified constraint (11) using rules (I), (II) which results in the following new constraint:
(13)    (LOOP(s(s(x33)), s(0), x11)≥IF(le(s(s(x33)), s(0)), s(s(x33)), s(0), x11))

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

IF(false, x, y, z) → LOOP(x, double(y), s(z))
- (IF(false, x3, s(x18), x5)≥LOOP(x3, double(s(x18)), s(x5)))
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)
- (LOOP(s(x30), s(x31), x11)≥IF(le(s(x30), s(x31)), s(x30), s(x31), x11) ⇒ LOOP(s(s(x30)), s(s(x31)), x11)≥IF(le(s(s(x30)), s(s(x31))), s(s(x30)), s(s(x31)), x11))
- (LOOP(s(s(x33)), s(0), x11)≥IF(le(s(s(x33)), s(0)), s(s(x33)), s(0), 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(IF(x₁, x₂, x₃, x₄)) = -1 - x₁ + x₂ - x₃
POL(LOOP(x₁, x₂, x₃)) = -1 + x₁ - x₂
POL(c) = -5
POL(double(x₁)) = 2·x₁
POL(false) = 2
POL(le(x₁, x₂)) = 1
POL(s(x₁)) = 2 + x₁
POL(true) = 1

The following pairs are in P_>:

LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)

The following pairs are in P_bound:

LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)

The following rules are usable:

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

IF(false, x, y, z) → LOOP(x, double(y), s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))

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.