Termination w.r.t. Q proof of /tmp/tmprQDPEM/thiemann20.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:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → 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:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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:

AVER(sum, z) → GT(sum, double(z))
AVERAGE(x, y) → AVER(plus(x, y), 0)
DOUBLE(s(x)) → DOUBLE(x)
GT(s(x), s(y)) → GT(x, y)
IF(true, sum, z) → AVER(sum, s(z))
AVER(sum, z) → IF(gt(sum, double(z)), sum, z)
AVER(sum, z) → DOUBLE(z)
AVERAGE(x, y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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:

AVER(sum, z) → GT(sum, double(z))
AVERAGE(x, y) → AVER(plus(x, y), 0)
DOUBLE(s(x)) → DOUBLE(x)
GT(s(x), s(y)) → GT(x, y)
IF(true, sum, z) → AVER(sum, s(z))
AVER(sum, z) → IF(gt(sum, double(z)), sum, z)
AVER(sum, z) → DOUBLE(z)
AVERAGE(x, y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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 4 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ 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:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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

              ↳ 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:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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.

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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

              ↳ 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

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

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

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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.

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

From the DPs we obtained the following set of size-change graphs:

PLUS(s(x), 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

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

GT(s(x), s(y)) → GT(x, y)

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

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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.

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

GT(s(x), s(y)) → GT(x, y)

From the DPs we obtained the following set of size-change graphs:

GT(s(x), s(y)) → GT(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

IF(true, sum, z) → AVER(sum, s(z))
AVER(sum, z) → IF(gt(sum, double(z)), sum, z)

The TRS R consists of the following rules:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

IF(true, sum, z) → AVER(sum, s(z))
AVER(sum, z) → IF(gt(sum, double(z)), sum, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(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.

plus(0, x0)
plus(s(x0), x1)
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

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

AVER(sum, z) → IF(gt(sum, double(z)), sum, z)
IF(true, sum, z) → AVER(sum, s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(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(true, sum, z) → AVER(sum, s(z)) the following chains were created:

We consider the chain IF(true, sum, z) → AVER(sum, s(z)), AVER(sum, z) → IF(gt(sum, double(z)), sum, z) which results in the following constraint:
(1) (AVER(x2, s(x3))=AVER(x4, x5) ⇒ IF(true, x2, x3)≥AVER(x2, s(x3)))

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

For Pair AVER(sum, z) → IF(gt(sum, double(z)), sum, z) the following chains were created:

We consider the chain AVER(sum, z) → IF(gt(sum, double(z)), sum, z), IF(true, sum, z) → AVER(sum, s(z)) which results in the following constraint:
(3)    (IF(gt(x6, double(x7)), x6, x7)=IF(true, x8, x9) ⇒ AVER(x6, x7)≥IF(gt(x6, double(x7)), x6, x7))

We simplified constraint (3) using rules (I), (II), (IV), (VII) which results in the following new constraint:
(4)    (double(x7)=x12∧gt(x6, x12)=true ⇒ AVER(x6, x7)≥IF(gt(x6, double(x7)), x6, x7))

We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on gt(x6, x12)=true which results in the following new constraints:
(5)    (true=true∧double(x7)=0 ⇒ AVER(s(x14), x7)≥IF(gt(s(x14), double(x7)), s(x14), x7))

(6)    (gt(x15, x16)=true∧double(x7)=s(x16)∧(∀x17:gt(x15, x16)=true∧double(x17)=x16 ⇒ AVER(x15, x17)≥IF(gt(x15, double(x17)), x15, x17)) ⇒ AVER(s(x15), x7)≥IF(gt(s(x15), double(x7)), s(x15), x7))

We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
(7)    (double(x7)=0 ⇒ AVER(s(x14), x7)≥IF(gt(s(x14), double(x7)), s(x14), x7))

We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on double(x7)=s(x16) which results in the following new constraint:
(8)    (s(s(double(x19)))=s(x16)∧gt(x15, x16)=true∧(∀x17:gt(x15, x16)=true∧double(x17)=x16 ⇒ AVER(x15, x17)≥IF(gt(x15, double(x17)), x15, x17))∧(∀x20,x21,x22:double(x19)=s(x20)∧gt(x21, x20)=true∧(∀x22:gt(x21, x20)=true∧double(x22)=x20 ⇒ AVER(x21, x22)≥IF(gt(x21, double(x22)), x21, x22)) ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) ⇒ AVER(s(x15), s(x19))≥IF(gt(s(x15), double(s(x19))), s(x15), s(x19)))

We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on double(x7)=0 which results in the following new constraint:
(9)    (0=0 ⇒ AVER(s(x14), 0)≥IF(gt(s(x14), double(0)), s(x14), 0))

We simplified constraint (9) using rules (I), (II) which results in the following new constraint:
(10)    (AVER(s(x14), 0)≥IF(gt(s(x14), double(0)), s(x14), 0))

We simplified constraint (8) using rules (I), (II), (IV) which results in the following new constraint:
(11)    (s(double(x19))=x16∧gt(x15, x16)=true∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) ⇒ AVER(s(x15), s(x19))≥IF(gt(s(x15), double(s(x19))), s(x15), s(x19)))

We simplified constraint (11) using rule (V) (with possible (I) afterwards) using induction on gt(x15, x16)=true which results in the following new constraints:
(12)    (true=true∧s(double(x19))=0∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) ⇒ AVER(s(s(x24)), s(x19))≥IF(gt(s(s(x24)), double(s(x19))), s(s(x24)), s(x19)))

(13)    (gt(x25, x26)=true∧s(double(x19))=s(x26)∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19))∧(∀x27,x28,x29:gt(x25, x26)=true∧s(double(x27))=x26∧(∀x28,x29:double(x27)=s(x28)∧gt(x29, x28)=true ⇒ AVER(s(x29), x27)≥IF(gt(s(x29), double(x27)), s(x29), x27)) ⇒ AVER(s(x25), s(x27))≥IF(gt(s(x25), double(s(x27))), s(x25), s(x27))) ⇒ AVER(s(s(x25)), s(x19))≥IF(gt(s(s(x25)), double(s(x19))), s(s(x25)), s(x19)))

We solved constraint (12) using rules (I), (II).We simplified constraint (13) using rules (I), (II), (IV) which results in the following new constraint:
(14)    (gt(x25, x26)=true∧double(x19)=x26∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) ⇒ AVER(s(s(x25)), s(x19))≥IF(gt(s(s(x25)), double(s(x19))), s(s(x25)), s(x19)))

We simplified constraint (14) using rule (V) (with possible (I) afterwards) using induction on gt(x25, x26)=true which results in the following new constraints:
(15)    (true=true∧double(x19)=0∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) ⇒ AVER(s(s(s(x31))), s(x19))≥IF(gt(s(s(s(x31))), double(s(x19))), s(s(s(x31))), s(x19)))

(16)    (gt(x32, x33)=true∧double(x19)=s(x33)∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19))∧(∀x34,x35,x36:gt(x32, x33)=true∧double(x34)=x33∧(∀x35,x36:double(x34)=s(x35)∧gt(x36, x35)=true ⇒ AVER(s(x36), x34)≥IF(gt(s(x36), double(x34)), s(x36), x34)) ⇒ AVER(s(s(x32)), s(x34))≥IF(gt(s(s(x32)), double(s(x34))), s(s(x32)), s(x34))) ⇒ AVER(s(s(s(x32))), s(x19))≥IF(gt(s(s(s(x32))), double(s(x19))), s(s(s(x32))), s(x19)))

We simplified constraint (15) using rules (I), (II), (IV) which results in the following new constraint:
(17)    (double(x19)=0 ⇒ AVER(s(s(s(x31))), s(x19))≥IF(gt(s(s(s(x31))), double(s(x19))), s(s(s(x31))), s(x19)))

We simplified constraint (16) using rule (VI) where we applied the induction hypothesis (∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) with σ = [x20 / x33, x21 / x32] which results in the following new constraint:
(18)    (AVER(s(x32), x19)≥IF(gt(s(x32), double(x19)), s(x32), x19)∧(∀x34,x35,x36:gt(x32, x33)=true∧double(x34)=x33∧(∀x35,x36:double(x34)=s(x35)∧gt(x36, x35)=true ⇒ AVER(s(x36), x34)≥IF(gt(s(x36), double(x34)), s(x36), x34)) ⇒ AVER(s(s(x32)), s(x34))≥IF(gt(s(s(x32)), double(s(x34))), s(s(x32)), s(x34))) ⇒ AVER(s(s(s(x32))), s(x19))≥IF(gt(s(s(s(x32))), double(s(x19))), s(s(s(x32))), s(x19)))

We simplified constraint (17) using rule (V) (with possible (I) afterwards) using induction on double(x19)=0 which results in the following new constraint:
(19)    (0=0 ⇒ AVER(s(s(s(x31))), s(0))≥IF(gt(s(s(s(x31))), double(s(0))), s(s(s(x31))), s(0)))

We simplified constraint (19) using rules (I), (II) which results in the following new constraint:
(20)    (AVER(s(s(s(x31))), s(0))≥IF(gt(s(s(s(x31))), double(s(0))), s(s(s(x31))), s(0)))

We simplified constraint (18) using rule (IV) which results in the following new constraint:
(21)    (AVER(s(x32), x19)≥IF(gt(s(x32), double(x19)), s(x32), x19) ⇒ AVER(s(s(s(x32))), s(x19))≥IF(gt(s(s(s(x32))), double(s(x19))), s(s(s(x32))), s(x19)))

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

IF(true, sum, z) → AVER(sum, s(z))
- (IF(true, x2, x3)≥AVER(x2, s(x3)))
AVER(sum, z) → IF(gt(sum, double(z)), sum, z)
- (AVER(s(x14), 0)≥IF(gt(s(x14), double(0)), s(x14), 0))
- (AVER(s(s(s(x31))), s(0))≥IF(gt(s(s(s(x31))), double(s(0))), s(s(s(x31))), s(0)))
- (AVER(s(x32), x19)≥IF(gt(s(x32), double(x19)), s(x32), x19) ⇒ AVER(s(s(s(x32))), s(x19))≥IF(gt(s(s(s(x32))), double(s(x19))), s(s(s(x32))), s(x19)))

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(AVER(x₁, x₂)) = -1 + x₁ - x₂
POL(IF(x₁, x₂, x₃)) = -1 - x₁ + x₂ - x₃
POL(c) = -1
POL(double(x₁)) = 0
POL(false) = 0
POL(gt(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following pairs are in P_>:

IF(true, sum, z) → AVER(sum, s(z))

The following pairs are in P_bound:

AVER(sum, z) → IF(gt(sum, double(z)), sum, z)

The following rules are usable:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ AND

                            ↳ QDP

                              ↳ DependencyGraphProof

                            ↳ QDP

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

AVER(sum, z) → IF(gt(sum, double(z)), sum, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(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.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ NonInfProof

                          ↳ AND

                            ↳ QDP

                            ↳ QDP

                              ↳ DependencyGraphProof

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

IF(true, sum, z) → AVER(sum, s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(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.