Termination proof of /tmp/tmpxAYKhS/divMinus.itrs

Termination of the given ITRSProblem could successfully be proven:

↳ ITRS

  ↳ ITRStoIDPProof

ITRS problem:
The following domains are used:

z

The TRS R consists of the following rules:

if(TRUE, x, y) → +@z(div(-@z(x, y), y), 1@z)
div(x, y) → if(&&(>=@z(x, y), >@z(y, 0@z)), x, y)

The set Q consists of the following terms:

if(TRUE, x0, x1)
div(x0, x1)

Added dependency pairs

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

if(TRUE, x, y) → +@z(div(-@z(x, y), y), 1@z)
div(x, y) → if(&&(>=@z(x, y), >@z(y, 0@z)), x, y)

The integer pair graph contains the following rules and edges:

(0): DIV(x[0], y[0]) → IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])
(1): IF(TRUE, x[1], y[1]) → DIV(-@z(x[1], y[1]), y[1])

(0) -> (1), if ((x[0] →^* x[1])∧(y[0] →^* y[1])∧(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)) →^* TRUE))

(1) -> (0), if ((y[1] →^* y[0])∧(-@z(x[1], y[1]) →^* x[0]))

The set Q consists of the following terms:

if(TRUE, x0, x1)
div(x0, x1)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(0): DIV(x[0], y[0]) → IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])
(1): IF(TRUE, x[1], y[1]) → DIV(-@z(x[1], y[1]), y[1])

(0) -> (1), if ((x[0] →^* x[1])∧(y[0] →^* y[1])∧(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)) →^* TRUE))

(1) -> (0), if ((y[1] →^* y[0])∧(-@z(x[1], y[1]) →^* x[0]))

The set Q consists of the following terms:

if(TRUE, x0, x1)
div(x0, x1)

The constraints were generated the following way:
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.

For Pair DIV(x, y) → IF(&&(>=@z(x, y), >@z(y, 0@z)), x, y) the following chains were created:

We consider the chain DIV(x[0], y[0]) → IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0]) which results in the following constraint:
(1)    (DIV(x[0], y[0])≥NonInfC∧DIV(x[0], y[0])≥IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])∧(U^Increasing(IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥))

We simplified constraint (1) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(2)    ((U^Increasing(IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (2) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(3)    ((U^Increasing(IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (3) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(4)    (0 ≥ 0∧0 ≥ 0∧(U^Increasing(IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥))

We simplified constraint (4) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(5)    (0 = 0∧0 = 0∧(U^Increasing(IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0∧0 = 0∧0 = 0)

For Pair IF(TRUE, x, y) → DIV(-@z(x, y), y) the following chains were created:

We consider the chain DIV(x[0], y[0]) → IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0]), IF(TRUE, x[1], y[1]) → DIV(-@z(x[1], y[1]), y[1]), DIV(x[0], y[0]) → IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0]) which results in the following constraint:
(6)    (y[1]=y[0]1∧-@z(x[1], y[1])=x[0]1∧y[0]=y[1]∧&&(>=@z(x[0], y[0]), >@z(y[0], 0@z))=TRUE∧x[0]=x[1] ⇒ IF(TRUE, x[1], y[1])≥NonInfC∧IF(TRUE, x[1], y[1])≥DIV(-@z(x[1], y[1]), y[1])∧(U^Increasing(DIV(-@z(x[1], y[1]), y[1])), ≥))

We simplified constraint (6) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:
(7)    (>=@z(x[0], y[0])=TRUE∧>@z(y[0], 0@z)=TRUE ⇒ IF(TRUE, x[0], y[0])≥NonInfC∧IF(TRUE, x[0], y[0])≥DIV(-@z(x[0], y[0]), y[0])∧(U^Increasing(DIV(-@z(x[1], y[1]), y[1])), ≥))

We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(8)    (x[0] + (-1)y[0] ≥ 0∧-1 + y[0] ≥ 0 ⇒ (U^Increasing(DIV(-@z(x[1], y[1]), y[1])), ≥)∧-1 + (-1)Bound + (-1)y[0] + (2)x[0] ≥ 0∧-2 + (2)y[0] ≥ 0)

We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(9)    (x[0] + (-1)y[0] ≥ 0∧-1 + y[0] ≥ 0 ⇒ (U^Increasing(DIV(-@z(x[1], y[1]), y[1])), ≥)∧-1 + (-1)Bound + (-1)y[0] + (2)x[0] ≥ 0∧-2 + (2)y[0] ≥ 0)

We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(10)    (x[0] + (-1)y[0] ≥ 0∧-1 + y[0] ≥ 0 ⇒ -2 + (2)y[0] ≥ 0∧-1 + (-1)Bound + (-1)y[0] + (2)x[0] ≥ 0∧(U^Increasing(DIV(-@z(x[1], y[1]), y[1])), ≥))

We simplified constraint (10) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(11)    (x[0] + -1 + (-1)y[0] ≥ 0∧y[0] ≥ 0 ⇒ (2)y[0] ≥ 0∧-2 + (-1)Bound + (-1)y[0] + (2)x[0] ≥ 0∧(U^Increasing(DIV(-@z(x[1], y[1]), y[1])), ≥))

We simplified constraint (11) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(12)    (x[0] ≥ 0∧y[0] ≥ 0 ⇒ (2)y[0] ≥ 0∧(-1)Bound + y[0] + (2)x[0] ≥ 0∧(U^Increasing(DIV(-@z(x[1], y[1]), y[1])), ≥))

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

DIV(x, y) → IF(&&(>=@z(x, y), >@z(y, 0@z)), x, y)
- (0 = 0∧0 = 0∧(U^Increasing(IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0∧0 = 0∧0 = 0)
IF(TRUE, x, y) → DIV(-@z(x, y), y)
- (x[0] ≥ 0∧y[0] ≥ 0 ⇒ (2)y[0] ≥ 0∧(-1)Bound + y[0] + (2)x[0] ≥ 0∧(U^Increasing(DIV(-@z(x[1], y[1]), y[1])), ≥))

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 over integers[POLO]:

POL(-@z(x₁, x₂)) = x₁ + (-1)x₂
POL(>=@z(x₁, x₂)) = -1
POL(0@z) = 0
POL(TRUE) = -1
POL(&&(x₁, x₂)) = 2
POL(DIV(x₁, x₂)) = (-1)x₂ + (2)x₁
POL(FALSE) = 2
POL(IF(x₁, x₂, x₃)) = -1 + (-1)x₃ + (2)x₂
POL(undefined) = -1
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

DIV(x[0], y[0]) → IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])
IF(TRUE, x[1], y[1]) → DIV(-@z(x[1], y[1]), y[1])

The following pairs are in P_bound:

IF(TRUE, x[1], y[1]) → DIV(-@z(x[1], y[1]), y[1])

The following pairs are in P_≥:
none

At least the following rules have been oriented under context sensitive arithmetic replacement:

&&(FALSE, FALSE)¹ ↔ FALSE¹
-@z¹ ↔
&&(TRUE, TRUE)¹ → TRUE¹
&&(FALSE, TRUE)¹ ↔ FALSE¹

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ AND

              ↳ IDP

                ↳ IDependencyGraphProof

              ↳ IDP

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(0): DIV(x[0], y[0]) → IF(&&(>=@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])

The set Q consists of the following terms:

if(TRUE, x0, x1)
div(x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ AND

              ↳ IDP

              ↳ IDP

                ↳ IDependencyGraphProof

I DP problem:
The following domains are used:none

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

if(TRUE, x0, x1)
div(x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs.