Termination proof of /tmp/tmpfjhE2l/div.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:

div(x, y) → Cond_div1(>=@z(y, x), x, y)
Cond_div1(TRUE, x, y) → 0@z
div(x, y) → Cond_div(&&(>@z(x, y), >@z(y, 0@z)), x, y)
Cond_div2(TRUE, x, y) → 0@z
div(x, y) → Cond_div2(>=@z(0@z, y), x, y)
Cond_div(TRUE, x, y) → +@z(div(-@z(x, y), y), 1@z)

The set Q consists of the following terms:

div(x0, x1)
Cond_div1(TRUE, x0, x1)
Cond_div2(TRUE, x0, x1)
Cond_div(TRUE, 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:

div(x, y) → Cond_div1(>=@z(y, x), x, y)
Cond_div1(TRUE, x, y) → 0@z
div(x, y) → Cond_div(&&(>@z(x, y), >@z(y, 0@z)), x, y)
Cond_div2(TRUE, x, y) → 0@z
div(x, y) → Cond_div2(>=@z(0@z, y), x, y)
Cond_div(TRUE, x, y) → +@z(div(-@z(x, y), y), 1@z)

The integer pair graph contains the following rules and edges:

(0): DIV(x[0], y[0]) → COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])
(1): DIV(x[1], y[1]) → COND_DIV2(>=@z(0@z, y[1]), x[1], y[1])
(2): COND_DIV(TRUE, x[2], y[2]) → DIV(-@z(x[2], y[2]), y[2])
(3): DIV(x[3], y[3]) → COND_DIV1(>=@z(y[3], x[3]), x[3], y[3])

(0) -> (2), if ((x[0] →^* x[2])∧(y[0] →^* y[2])∧(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)) →^* TRUE))

(2) -> (0), if ((y[2] →^* y[0])∧(-@z(x[2], y[2]) →^* x[0]))

(2) -> (1), if ((y[2] →^* y[1])∧(-@z(x[2], y[2]) →^* x[1]))

(2) -> (3), if ((y[2] →^* y[3])∧(-@z(x[2], y[2]) →^* x[3]))

The set Q consists of the following terms:

div(x0, x1)
Cond_div1(TRUE, x0, x1)
Cond_div2(TRUE, x0, x1)
Cond_div(TRUE, 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

          ↳ IDependencyGraphProof

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]) → COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])
(1): DIV(x[1], y[1]) → COND_DIV2(>=@z(0@z, y[1]), x[1], y[1])
(2): COND_DIV(TRUE, x[2], y[2]) → DIV(-@z(x[2], y[2]), y[2])
(3): DIV(x[3], y[3]) → COND_DIV1(>=@z(y[3], x[3]), x[3], y[3])

(0) -> (2), if ((x[0] →^* x[2])∧(y[0] →^* y[2])∧(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)) →^* TRUE))

(2) -> (0), if ((y[2] →^* y[0])∧(-@z(x[2], y[2]) →^* x[0]))

(2) -> (1), if ((y[2] →^* y[1])∧(-@z(x[2], y[2]) →^* x[1]))

(2) -> (3), if ((y[2] →^* y[3])∧(-@z(x[2], y[2]) →^* x[3]))

The set Q consists of the following terms:

div(x0, x1)
Cond_div1(TRUE, x0, x1)
Cond_div2(TRUE, x0, x1)
Cond_div(TRUE, x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ IDP

              ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

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

(2): COND_DIV(TRUE, x[2], y[2]) → DIV(-@z(x[2], y[2]), y[2])
(0): DIV(x[0], y[0]) → COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])

(2) -> (0), if ((y[2] →^* y[0])∧(-@z(x[2], y[2]) →^* x[0]))

(0) -> (2), if ((x[0] →^* x[2])∧(y[0] →^* y[2])∧(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)) →^* TRUE))

The set Q consists of the following terms:

div(x0, x1)
Cond_div1(TRUE, x0, x1)
Cond_div2(TRUE, x0, x1)
Cond_div(TRUE, 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 COND_DIV(TRUE, x[2], y[2]) → DIV(-@z(x[2], y[2]), y[2]) the following chains were created:

We consider the chain DIV(x[0], y[0]) → COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0]), COND_DIV(TRUE, x[2], y[2]) → DIV(-@z(x[2], y[2]), y[2]), DIV(x[0], y[0]) → COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0]) which results in the following constraint:
(1)    (x[0]=x[2]∧y[0]=y[2]∧y[2]=y[0]1∧&&(>@z(x[0], y[0]), >@z(y[0], 0@z))=TRUE∧-@z(x[2], y[2])=x[0]1 ⇒ COND_DIV(TRUE, x[2], y[2])≥NonInfC∧COND_DIV(TRUE, x[2], y[2])≥DIV(-@z(x[2], y[2]), y[2])∧(U^Increasing(DIV(-@z(x[2], y[2]), y[2])), ≥))

We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:
(2)    (>@z(x[0], y[0])=TRUE∧>@z(y[0], 0@z)=TRUE ⇒ COND_DIV(TRUE, x[0], y[0])≥NonInfC∧COND_DIV(TRUE, x[0], y[0])≥DIV(-@z(x[0], y[0]), y[0])∧(U^Increasing(DIV(-@z(x[2], y[2]), y[2])), ≥))

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

We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(4)    (x[0] + -1 + (-1)y[0] ≥ 0∧-1 + y[0] ≥ 0 ⇒ (U^Increasing(DIV(-@z(x[2], y[2]), y[2])), ≥)∧-1 + (-1)Bound + y[0] + (2)x[0] ≥ 0∧-2 + (2)y[0] ≥ 0)

We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5)    (-1 + y[0] ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0 ⇒ (U^Increasing(DIV(-@z(x[2], y[2]), y[2])), ≥)∧-2 + (2)y[0] ≥ 0∧-1 + (-1)Bound + y[0] + (2)x[0] ≥ 0)

We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(6)    (y[0] ≥ 0∧x[0] + -2 + (-1)y[0] ≥ 0 ⇒ (U^Increasing(DIV(-@z(x[2], y[2]), y[2])), ≥)∧(2)y[0] ≥ 0∧(-1)Bound + y[0] + (2)x[0] ≥ 0)

We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(7)    (y[0] ≥ 0∧x[0] ≥ 0 ⇒ (U^Increasing(DIV(-@z(x[2], y[2]), y[2])), ≥)∧(2)y[0] ≥ 0∧4 + (-1)Bound + (3)y[0] + (2)x[0] ≥ 0)

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

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

We simplified constraint (8) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(9)    ((U^Increasing(COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (9) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(10)    ((U^Increasing(COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (10) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(11)    (0 ≥ 0∧(U^Increasing(COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥)∧0 ≥ 0)

We simplified constraint (11) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(12)    (0 ≥ 0∧0 = 0∧(U^Increasing(COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥)∧0 ≥ 0∧0 = 0∧0 = 0∧0 = 0)

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

COND_DIV(TRUE, x[2], y[2]) → DIV(-@z(x[2], y[2]), y[2])
- (y[0] ≥ 0∧x[0] ≥ 0 ⇒ (U^Increasing(DIV(-@z(x[2], y[2]), y[2])), ≥)∧(2)y[0] ≥ 0∧4 + (-1)Bound + (3)y[0] + (2)x[0] ≥ 0)
DIV(x[0], y[0]) → COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])
- (0 ≥ 0∧0 = 0∧(U^Increasing(COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])), ≥)∧0 ≥ 0∧0 = 0∧0 = 0∧0 = 0)

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

The following pairs are in P_>:

COND_DIV(TRUE, x[2], y[2]) → DIV(-@z(x[2], y[2]), y[2])
DIV(x[0], y[0]) → COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])

The following pairs are in P_bound:

COND_DIV(TRUE, x[2], y[2]) → DIV(-@z(x[2], y[2]), y[2])

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, FALSE)¹ → FALSE¹
&&(FALSE, TRUE)¹ ↔ FALSE¹

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ 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]) → COND_DIV(&&(>@z(x[0], y[0]), >@z(y[0], 0@z)), x[0], y[0])

The set Q consists of the following terms:

div(x0, x1)
Cond_div1(TRUE, x0, x1)
Cond_div2(TRUE, x0, x1)
Cond_div(TRUE, x0, x1)

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

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ 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:

div(x0, x1)
Cond_div1(TRUE, x0, x1)
Cond_div2(TRUE, x0, x1)
Cond_div(TRUE, x0, x1)

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