Termination proof of /tmp/tmpZd0bJ4/mult.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:

Cond_mult1(TRUE, x, y) → -@z(mult(-@z(x), y))
mult(0@z, y) → 0@z
mult(x, y) → Cond_mult(>@z(x, 0@z), x, y)
mult(x, y) → Cond_mult1(>@z(0@z, x), x, y)
Cond_mult(TRUE, x, y) → +@z(mult(-@z(x, 1@z), y), y)

The set Q consists of the following terms:

Cond_mult1(TRUE, x0, x1)
mult(x0, x1)
Cond_mult(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:

Cond_mult1(TRUE, x, y) → -@z(mult(-@z(x), y))
mult(0@z, y) → 0@z
mult(x, y) → Cond_mult(>@z(x, 0@z), x, y)
mult(x, y) → Cond_mult1(>@z(0@z, x), x, y)
Cond_mult(TRUE, x, y) → +@z(mult(-@z(x, 1@z), y), y)

The integer pair graph contains the following rules and edges:

(0): MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0])
(1): COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1])
(2): COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2])
(3): MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3])

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

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

(1) -> (3), if ((y[1] →^* y[3])∧(-@z(x[1], 1@z) →^* x[3]))

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

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

(3) -> (1), if ((x[3] →^* x[1])∧(y[3] →^* y[1])∧(>@z(x[3], 0@z) →^* TRUE))

The set Q consists of the following terms:

Cond_mult1(TRUE, x0, x1)
mult(x0, x1)
Cond_mult(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

          ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

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

(0): MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0])
(1): COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1])
(2): COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2])
(3): MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3])

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

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

(1) -> (3), if ((y[1] →^* y[3])∧(-@z(x[1], 1@z) →^* x[3]))

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

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

(3) -> (1), if ((x[3] →^* x[1])∧(y[3] →^* y[1])∧(>@z(x[3], 0@z) →^* TRUE))

The set Q consists of the following terms:

Cond_mult1(TRUE, x0, x1)
mult(x0, x1)
Cond_mult(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 MULT(x, y) → COND_MULT1(>@z(0@z, x), x, y) the following chains were created:

We consider the chain MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]) which results in the following constraint:
(1)    (MULT(x[0], y[0])≥NonInfC∧MULT(x[0], y[0])≥COND_MULT1(>@z(0@z, x[0]), x[0], y[0])∧(U^Increasing(COND_MULT1(>@z(0@z, x[0]), x[0], y[0])), ≥))

We simplified constraint (1) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(2)    ((U^Increasing(COND_MULT1(>@z(0@z, x[0]), 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(COND_MULT1(>@z(0@z, x[0]), 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∧(U^Increasing(COND_MULT1(>@z(0@z, x[0]), x[0], y[0])), ≥)∧0 ≥ 0)

We simplified constraint (4) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(5)    (0 ≥ 0∧0 = 0∧0 = 0∧0 = 0∧(U^Increasing(COND_MULT1(>@z(0@z, x[0]), x[0], y[0])), ≥)∧0 = 0∧0 ≥ 0)

For Pair COND_MULT(TRUE, x, y) → MULT(-@z(x, 1@z), y) the following chains were created:

We consider the chain MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]), MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]), MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]) which results in the following constraint:
(6)    (x[0]=x[2]∧x[0]1=x[2]1∧>@z(0@z, x[0]1)=TRUE∧y[0]1=y[2]1∧y[1]=y[0]1∧-@z(x[2])=x[3]∧y[0]=y[2]∧x[3]=x[1]∧y[3]=y[1]∧y[2]=y[3]∧>@z(x[3], 0@z)=TRUE∧>@z(0@z, x[0])=TRUE∧-@z(x[1], 1@z)=x[0]1 ⇒ COND_MULT(TRUE, x[1], y[1])≥NonInfC∧COND_MULT(TRUE, x[1], y[1])≥MULT(-@z(x[1], 1@z), y[1])∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))

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

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

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

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

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

We solved constraint (11) using rule (IDP_SMT_SPLIT).
We consider the chain MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]), MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]), MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]) which results in the following constraint:
(12)    (-@z(x[1]1, 1@z)=x[3]2∧y[1]1=y[3]2∧y[1]=y[3]1∧x[3]=x[1]∧>@z(x[3]2, 0@z)=TRUE∧x[3]1=x[1]1∧y[3]=y[1]∧>@z(x[3], 0@z)=TRUE∧x[3]2=x[1]2∧-@z(x[1], 1@z)=x[3]1∧y[3]2=y[1]2∧>@z(x[3]1, 0@z)=TRUE∧y[3]1=y[1]1 ⇒ COND_MULT(TRUE, x[1]1, y[1]1)≥NonInfC∧COND_MULT(TRUE, x[1]1, y[1]1)≥MULT(-@z(x[1]1, 1@z), y[1]1)∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥))

We simplified constraint (12) using rules (III), (IV) which results in the following new constraint:
(13)    (>@z(-@z(-@z(x[3], 1@z), 1@z), 0@z)=TRUE∧>@z(x[3], 0@z)=TRUE∧>@z(-@z(x[3], 1@z), 0@z)=TRUE ⇒ COND_MULT(TRUE, -@z(x[3], 1@z), y[3])≥NonInfC∧COND_MULT(TRUE, -@z(x[3], 1@z), y[3])≥MULT(-@z(-@z(x[3], 1@z), 1@z), y[3])∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥))

We simplified constraint (13) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(14)    (-3 + x[3] ≥ 0∧-1 + x[3] ≥ 0∧-2 + x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (14) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(15)    (-3 + x[3] ≥ 0∧-1 + x[3] ≥ 0∧-2 + x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (15) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(16)    (-2 + x[3] ≥ 0∧-3 + x[3] ≥ 0∧-1 + x[3] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 ≥ 0)

We simplified constraint (16) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(17)    (-2 + x[3] ≥ 0∧-3 + x[3] ≥ 0∧-1 + x[3] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 = 0∧0 = 0)

We simplified constraint (17) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(18)    (x[3] ≥ 0∧-1 + x[3] ≥ 0∧1 + x[3] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 = 0∧0 = 0)

We simplified constraint (18) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(19)    (1 + x[3] ≥ 0∧x[3] ≥ 0∧2 + x[3] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 = 0∧0 = 0)
We consider the chain MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]), MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]), MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]) which results in the following constraint:
(20)    (x[0]=x[2]∧y[1]=y[3]1∧y[1]1=y[0]∧y[0]=y[2]∧x[3]=x[1]∧x[3]1=x[1]1∧y[3]=y[1]∧>@z(x[3], 0@z)=TRUE∧>@z(0@z, x[0])=TRUE∧-@z(x[1], 1@z)=x[3]1∧-@z(x[1]1, 1@z)=x[0]∧>@z(x[3]1, 0@z)=TRUE∧y[3]1=y[1]1 ⇒ COND_MULT(TRUE, x[1]1, y[1]1)≥NonInfC∧COND_MULT(TRUE, x[1]1, y[1]1)≥MULT(-@z(x[1]1, 1@z), y[1]1)∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥))

We simplified constraint (20) using rules (III), (IV) which results in the following new constraint:
(21)    (>@z(x[3], 0@z)=TRUE∧>@z(0@z, -@z(-@z(x[3], 1@z), 1@z))=TRUE∧>@z(-@z(x[3], 1@z), 0@z)=TRUE ⇒ COND_MULT(TRUE, -@z(x[3], 1@z), y[3])≥NonInfC∧COND_MULT(TRUE, -@z(x[3], 1@z), y[3])≥MULT(-@z(-@z(x[3], 1@z), 1@z), y[3])∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥))

We simplified constraint (21) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(22)    (-1 + x[3] ≥ 0∧1 + (-1)x[3] ≥ 0∧-2 + x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (22) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(23)    (-1 + x[3] ≥ 0∧1 + (-1)x[3] ≥ 0∧-2 + x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (23) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(24)    (-2 + x[3] ≥ 0∧1 + (-1)x[3] ≥ 0∧-1 + x[3] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 ≥ 0)

We simplified constraint (24) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(25)    (-2 + x[3] ≥ 0∧1 + (-1)x[3] ≥ 0∧-1 + x[3] ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 = 0∧0 ≥ 0)

We solved constraint (25) using rule (IDP_SMT_SPLIT).
We consider the chain MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]), MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]), MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]) which results in the following constraint:
(26)    (x[0]=x[2]∧y[1]=y[3]1∧-@z(x[2])=x[3]∧y[0]=y[2]∧x[3]=x[1]∧x[3]1=x[1]1∧y[3]=y[1]∧y[2]=y[3]∧>@z(x[3], 0@z)=TRUE∧>@z(0@z, x[0])=TRUE∧-@z(x[1], 1@z)=x[3]1∧>@z(x[3]1, 0@z)=TRUE∧y[3]1=y[1]1 ⇒ COND_MULT(TRUE, x[1], y[1])≥NonInfC∧COND_MULT(TRUE, x[1], y[1])≥MULT(-@z(x[1], 1@z), y[1])∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))

We simplified constraint (26) using rules (III), (IV) which results in the following new constraint:
(27)    (>@z(-@z(x[0]), 0@z)=TRUE∧>@z(0@z, x[0])=TRUE∧>@z(-@z(-@z(x[0]), 1@z), 0@z)=TRUE ⇒ COND_MULT(TRUE, -@z(x[0]), y[0])≥NonInfC∧COND_MULT(TRUE, -@z(x[0]), y[0])≥MULT(-@z(-@z(x[0]), 1@z), y[0])∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))

We simplified constraint (27) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(28)    (-1 + (-1)x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0∧-2 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (28) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(29)    (-1 + (-1)x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0∧-2 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (29) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(30)    (-2 + (-1)x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))

We simplified constraint (30) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(31)    (-2 + (-1)x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧0 = 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))

We simplified constraint (31) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(32)    (-1 + x[0] ≥ 0∧x[0] ≥ 0∧x[0] ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧0 = 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))

We simplified constraint (32) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(33)    (x[0] ≥ 0∧1 + x[0] ≥ 0∧1 + x[0] ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧0 = 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))

For Pair COND_MULT1(TRUE, x, y) → MULT(-@z(x), y) the following chains were created:

We consider the chain MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]), MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]), MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]) which results in the following constraint:
(34)    (x[0]=x[2]∧x[0]1=x[2]1∧>@z(0@z, x[0]1)=TRUE∧y[0]1=y[2]1∧-@z(x[2])=x[0]1∧-@z(x[1], 1@z)=x[0]∧y[0]=y[2]∧x[3]=x[1]∧y[3]=y[1]∧>@z(x[3], 0@z)=TRUE∧>@z(0@z, x[0])=TRUE∧y[2]=y[0]1∧y[1]=y[0] ⇒ COND_MULT1(TRUE, x[2], y[2])≥NonInfC∧COND_MULT1(TRUE, x[2], y[2])≥MULT(-@z(x[2]), y[2])∧(U^Increasing(MULT(-@z(x[2]), y[2])), ≥))

We simplified constraint (34) using rules (III), (IV) which results in the following new constraint:
(35)    (>@z(0@z, -@z(-@z(x[3], 1@z)))=TRUE∧>@z(x[3], 0@z)=TRUE∧>@z(0@z, -@z(x[3], 1@z))=TRUE ⇒ COND_MULT1(TRUE, -@z(x[3], 1@z), y[3])≥NonInfC∧COND_MULT1(TRUE, -@z(x[3], 1@z), y[3])≥MULT(-@z(-@z(x[3], 1@z)), y[3])∧(U^Increasing(MULT(-@z(x[2]), y[2])), ≥))

We simplified constraint (35) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(36)    (-2 + x[3] ≥ 0∧-1 + x[3] ≥ 0∧(-1)x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]), y[2])), ≥)∧-2 + (-1)Bound + (-1)y[3] + x[3] ≥ 0∧-2 + (2)x[3] ≥ 0)

We simplified constraint (36) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(37)    (-2 + x[3] ≥ 0∧-1 + x[3] ≥ 0∧(-1)x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]), y[2])), ≥)∧-2 + (-1)Bound + (-1)y[3] + x[3] ≥ 0∧-2 + (2)x[3] ≥ 0)

We simplified constraint (37) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(38)    (-2 + x[3] ≥ 0∧-1 + x[3] ≥ 0∧(-1)x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]), y[2])), ≥)∧-2 + (-1)Bound + (-1)y[3] + x[3] ≥ 0∧-2 + (2)x[3] ≥ 0)

We simplified constraint (38) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(39)    (-2 + x[3] ≥ 0∧-1 + x[3] ≥ 0∧(-1)x[3] ≥ 0 ⇒ -1 = 0∧-2 + (2)x[3] ≥ 0∧(U^Increasing(MULT(-@z(x[2]), y[2])), ≥)∧0 = 0∧-2 + (-1)Bound + x[3] ≥ 0)

We solved constraint (39) using rule (IDP_SMT_SPLIT).
We consider the chain MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]), MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]), MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]) which results in the following constraint:
(40)    (x[0]=x[2]∧x[0]1=x[2]1∧>@z(0@z, x[0]1)=TRUE∧y[0]1=y[2]1∧-@z(x[2]1)=x[3]∧y[2]1=y[3]∧-@z(x[2])=x[0]1∧y[0]=y[2]∧x[3]=x[1]∧y[3]=y[1]∧>@z(x[3], 0@z)=TRUE∧>@z(0@z, x[0])=TRUE∧y[2]=y[0]1 ⇒ COND_MULT1(TRUE, x[2]1, y[2]1)≥NonInfC∧COND_MULT1(TRUE, x[2]1, y[2]1)≥MULT(-@z(x[2]1), y[2]1)∧(U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥))

We simplified constraint (40) using rules (III), (IV) which results in the following new constraint:
(41)    (>@z(0@z, -@z(x[0]))=TRUE∧>@z(-@z(-@z(x[0])), 0@z)=TRUE∧>@z(0@z, x[0])=TRUE ⇒ COND_MULT1(TRUE, -@z(x[0]), y[0])≥NonInfC∧COND_MULT1(TRUE, -@z(x[0]), y[0])≥MULT(-@z(-@z(x[0])), y[0])∧(U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥))

We simplified constraint (41) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(42)    (-1 + x[0] ≥ 0∧-1 + x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥)∧-1 + (-1)Bound + (-1)y[0] + (-1)x[0] ≥ 0∧(-2)x[0] ≥ 0)

We simplified constraint (42) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(43)    (-1 + x[0] ≥ 0∧-1 + x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥)∧-1 + (-1)Bound + (-1)y[0] + (-1)x[0] ≥ 0∧(-2)x[0] ≥ 0)

We simplified constraint (43) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(44)    (-1 + x[0] ≥ 0∧-1 + x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ -1 + (-1)Bound + (-1)y[0] + (-1)x[0] ≥ 0∧(U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥)∧(-2)x[0] ≥ 0)

We simplified constraint (44) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(45)    (-1 + x[0] ≥ 0∧-1 + x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥)∧-1 + (-1)Bound + (-1)x[0] ≥ 0∧0 = 0∧(-2)x[0] ≥ 0∧-1 = 0)

We solved constraint (45) using rule (IDP_SMT_SPLIT).
We consider the chain MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]), MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]), MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]) which results in the following constraint:
(46)    (x[0]=x[2]∧-@z(x[1], 1@z)=x[0]∧y[0]=y[2]∧x[3]=x[1]∧x[3]1=x[1]1∧y[3]=y[1]∧y[2]=y[3]1∧>@z(x[3], 0@z)=TRUE∧>@z(0@z, x[0])=TRUE∧y[1]=y[0]∧-@z(x[2])=x[3]1∧>@z(x[3]1, 0@z)=TRUE∧y[3]1=y[1]1 ⇒ COND_MULT1(TRUE, x[2], y[2])≥NonInfC∧COND_MULT1(TRUE, x[2], y[2])≥MULT(-@z(x[2]), y[2])∧(U^Increasing(MULT(-@z(x[2]), y[2])), ≥))

We simplified constraint (46) using rules (III), (IV) which results in the following new constraint:
(47)    (>@z(x[3], 0@z)=TRUE∧>@z(0@z, -@z(x[3], 1@z))=TRUE∧>@z(-@z(-@z(x[3], 1@z)), 0@z)=TRUE ⇒ COND_MULT1(TRUE, -@z(x[3], 1@z), y[3])≥NonInfC∧COND_MULT1(TRUE, -@z(x[3], 1@z), y[3])≥MULT(-@z(-@z(x[3], 1@z)), y[3])∧(U^Increasing(MULT(-@z(x[2]), y[2])), ≥))

We simplified constraint (47) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(48)    (-1 + x[3] ≥ 0∧(-1)x[3] ≥ 0∧(-1)x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]), y[2])), ≥)∧-2 + (-1)Bound + (-1)y[3] + x[3] ≥ 0∧-2 + (2)x[3] ≥ 0)

We simplified constraint (48) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(49)    (-1 + x[3] ≥ 0∧(-1)x[3] ≥ 0∧(-1)x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]), y[2])), ≥)∧-2 + (-1)Bound + (-1)y[3] + x[3] ≥ 0∧-2 + (2)x[3] ≥ 0)

We simplified constraint (49) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(50)    (-1 + x[3] ≥ 0∧(-1)x[3] ≥ 0∧(-1)x[3] ≥ 0 ⇒ -2 + (-1)Bound + (-1)y[3] + x[3] ≥ 0∧(U^Increasing(MULT(-@z(x[2]), y[2])), ≥)∧-2 + (2)x[3] ≥ 0)

We simplified constraint (50) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(51)    (-1 + x[3] ≥ 0∧(-1)x[3] ≥ 0∧(-1)x[3] ≥ 0 ⇒ 0 = 0∧-2 + (-1)Bound + x[3] ≥ 0∧-2 + (2)x[3] ≥ 0∧(U^Increasing(MULT(-@z(x[2]), y[2])), ≥)∧-1 = 0)

We solved constraint (51) using rule (IDP_SMT_SPLIT).
We consider the chain MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]), MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]), MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]) which results in the following constraint:
(52)    (x[0]=x[2]∧x[0]1=x[2]1∧>@z(0@z, x[0]1)=TRUE∧y[0]1=y[2]1∧y[0]2=y[2]2∧x[0]2=x[2]2∧-@z(x[2])=x[0]1∧y[2]1=y[0]2∧-@z(x[2]1)=x[0]2∧y[0]=y[2]∧>@z(0@z, x[0]2)=TRUE∧>@z(0@z, x[0])=TRUE∧y[2]=y[0]1 ⇒ COND_MULT1(TRUE, x[2]1, y[2]1)≥NonInfC∧COND_MULT1(TRUE, x[2]1, y[2]1)≥MULT(-@z(x[2]1), y[2]1)∧(U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥))

We simplified constraint (52) using rules (III), (IV) which results in the following new constraint:
(53)    (>@z(0@z, -@z(x[0]))=TRUE∧>@z(0@z, -@z(-@z(x[0])))=TRUE∧>@z(0@z, x[0])=TRUE ⇒ COND_MULT1(TRUE, -@z(x[0]), y[0])≥NonInfC∧COND_MULT1(TRUE, -@z(x[0]), y[0])≥MULT(-@z(-@z(x[0])), y[0])∧(U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥))

We simplified constraint (53) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(54)    (-1 + x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥)∧-1 + (-1)Bound + (-1)y[0] + (-1)x[0] ≥ 0∧(-2)x[0] ≥ 0)

We simplified constraint (54) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(55)    (-1 + x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥)∧-1 + (-1)Bound + (-1)y[0] + (-1)x[0] ≥ 0∧(-2)x[0] ≥ 0)

We simplified constraint (55) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(56)    (-1 + (-1)x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0∧-1 + x[0] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥)∧-1 + (-1)Bound + (-1)y[0] + (-1)x[0] ≥ 0∧(-2)x[0] ≥ 0)

We simplified constraint (56) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(57)    (-1 + (-1)x[0] ≥ 0∧-1 + (-1)x[0] ≥ 0∧-1 + x[0] ≥ 0 ⇒ 0 = 0∧(U^Increasing(MULT(-@z(x[2]1), y[2]1)), ≥)∧-1 = 0∧(-2)x[0] ≥ 0∧-1 + (-1)Bound + (-1)x[0] ≥ 0)

We solved constraint (57) using rule (IDP_SMT_SPLIT).

For Pair MULT(x, y) → COND_MULT(>@z(x, 0@z), x, y) the following chains were created:

We consider the chain MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]) which results in the following constraint:
(58)    (MULT(x[3], y[3])≥NonInfC∧MULT(x[3], y[3])≥COND_MULT(>@z(x[3], 0@z), x[3], y[3])∧(U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥))

We simplified constraint (58) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(59)    ((U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (59) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(60)    ((U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (60) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(61)    (0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥))

We simplified constraint (61) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(62)    (0 ≥ 0∧0 ≥ 0∧0 = 0∧0 = 0∧0 = 0∧(U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥)∧0 = 0)

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

MULT(x, y) → COND_MULT1(>@z(0@z, x), x, y)
- (0 ≥ 0∧0 = 0∧0 = 0∧0 = 0∧(U^Increasing(COND_MULT1(>@z(0@z, x[0]), x[0], y[0])), ≥)∧0 = 0∧0 ≥ 0)
COND_MULT(TRUE, x, y) → MULT(-@z(x, 1@z), y)
- (1 + x[3] ≥ 0∧x[3] ≥ 0∧2 + x[3] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1]1, 1@z), y[1]1)), ≥)∧0 = 0∧0 = 0)
- (x[0] ≥ 0∧1 + x[0] ≥ 0∧1 + x[0] ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧0 = 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))
COND_MULT1(TRUE, x, y) → MULT(-@z(x), y)
MULT(x, y) → COND_MULT(>@z(x, 0@z), x, y)
- (0 ≥ 0∧0 ≥ 0∧0 = 0∧0 = 0∧0 = 0∧(U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥)∧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(-@z(x₁)) = (-1)x₁
POL(COND_MULT(x₁, x₂, x₃)) = -1 + (-1)x₃ + x₂
POL(0@z) = 0
POL(MULT(x₁, x₂)) = -1 + (-1)x₂ + x₁
POL(TRUE) = -1
POL(COND_MULT1(x₁, x₂, x₃)) = -1 + (-1)x₃ + x₂
POL(FALSE) = -1
POL(1@z) = 1
POL(undefined) = -1
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1])

The following pairs are in P_bound:

COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2])

The following pairs are in P_≥:

MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0])
COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2])
MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3])

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

-@z¹ ↔

↳ 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): MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0])
(1): COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1])
(3): MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3])

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

(3) -> (1), if ((x[3] →^* x[1])∧(y[3] →^* y[1])∧(>@z(x[3], 0@z) →^* TRUE))

(1) -> (3), if ((y[1] →^* y[3])∧(-@z(x[1], 1@z) →^* x[3]))

The set Q consists of the following terms:

Cond_mult1(TRUE, x0, x1)
mult(x0, x1)
Cond_mult(TRUE, x0, x1)

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

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ AND

              ↳ IDP

                ↳ IDependencyGraphProof

                  ↳ IDP

                    ↳ IDPNonInfProof

              ↳ IDP

I DP problem:
The following domains are used:

z

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

(1): COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1])
(3): MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3])

(3) -> (1), if ((x[3] →^* x[1])∧(y[3] →^* y[1])∧(>@z(x[3], 0@z) →^* TRUE))

(1) -> (3), if ((y[1] →^* y[3])∧(-@z(x[1], 1@z) →^* x[3]))

The set Q consists of the following terms:

Cond_mult1(TRUE, x0, x1)
mult(x0, x1)
Cond_mult(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_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]) the following chains were created:

We consider the chain MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]), COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1]), MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]) which results in the following constraint:
(1)    (x[3]=x[1]∧y[3]=y[1]∧>@z(x[3], 0@z)=TRUE∧-@z(x[1], 1@z)=x[3]1∧y[1]=y[3]1 ⇒ COND_MULT(TRUE, x[1], y[1])≥NonInfC∧COND_MULT(TRUE, x[1], y[1])≥MULT(-@z(x[1], 1@z), y[1])∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))

We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
(2)    (>@z(x[3], 0@z)=TRUE ⇒ COND_MULT(TRUE, x[3], y[3])≥NonInfC∧COND_MULT(TRUE, x[3], y[3])≥MULT(-@z(x[3], 1@z), y[3])∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))

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

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

We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5)    (-1 + x[3] ≥ 0 ⇒ -1 + (-1)Bound + x[3] ≥ 0∧0 ≥ 0∧(U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥))

We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(6)    (-1 + x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥)∧0 ≥ 0∧0 = 0∧0 = 0∧-1 + (-1)Bound + x[3] ≥ 0)

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

For Pair MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]) the following chains were created:

We consider the chain MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3]) which results in the following constraint:
(8)    (MULT(x[3], y[3])≥NonInfC∧MULT(x[3], y[3])≥COND_MULT(>@z(x[3], 0@z), x[3], y[3])∧(U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥))

We simplified constraint (8) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(9)    ((U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (9) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(10)    ((U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (10) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(11)    (0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥))

We simplified constraint (11) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(12)    (0 ≥ 0∧0 = 0∧0 = 0∧0 = 0∧0 ≥ 0∧(U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥)∧0 = 0)

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

COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1])
- (x[3] ≥ 0 ⇒ (U^Increasing(MULT(-@z(x[1], 1@z), y[1])), ≥)∧0 ≥ 0∧0 = 0∧0 = 0∧(-1)Bound + x[3] ≥ 0)
MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3])
- (0 ≥ 0∧0 = 0∧0 = 0∧0 = 0∧0 ≥ 0∧(U^Increasing(COND_MULT(>@z(x[3], 0@z), x[3], y[3])), ≥)∧0 = 0)

POL(-@z(x₁, x₂)) = x₁ + (-1)x₂
POL(COND_MULT(x₁, x₂, x₃)) = -1 + x₂
POL(0@z) = 0
POL(MULT(x₁, x₂)) = -1 + x₁
POL(TRUE) = 0
POL(FALSE) = -1
POL(1@z) = 1
POL(undefined) = -1
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1])

The following pairs are in P_bound:

COND_MULT(TRUE, x[1], y[1]) → MULT(-@z(x[1], 1@z), y[1])

The following pairs are in P_≥:

MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3])

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

-@z¹ ↔

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ AND

              ↳ IDP

                ↳ IDependencyGraphProof

                  ↳ IDP

                    ↳ IDPNonInfProof

                      ↳ 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:

(3): MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3])

The set Q consists of the following terms:

Cond_mult1(TRUE, x0, x1)
mult(x0, x1)
Cond_mult(TRUE, 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:

z

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

(0): MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0])
(2): COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2])
(3): MULT(x[3], y[3]) → COND_MULT(>@z(x[3], 0@z), x[3], y[3])

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

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

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

The set Q consists of the following terms:

Cond_mult1(TRUE, x0, x1)
mult(x0, x1)
Cond_mult(TRUE, x0, x1)

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

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ AND

              ↳ IDP

              ↳ 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_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2])
(0): MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0])

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

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

The set Q consists of the following terms:

Cond_mult1(TRUE, x0, x1)
mult(x0, x1)
Cond_mult(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_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]) the following chains were created:

We consider the chain MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]), COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2]), MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), 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(0@z, x[0])=TRUE∧-@z(x[2])=x[0]1 ⇒ COND_MULT1(TRUE, x[2], y[2])≥NonInfC∧COND_MULT1(TRUE, x[2], y[2])≥MULT(-@z(x[2]), y[2])∧(U^Increasing(MULT(-@z(x[2]), y[2])), ≥))

We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
(2)    (>@z(0@z, x[0])=TRUE ⇒ COND_MULT1(TRUE, x[0], y[0])≥NonInfC∧COND_MULT1(TRUE, x[0], y[0])≥MULT(-@z(x[0]), y[0])∧(U^Increasing(MULT(-@z(x[2]), y[2])), ≥))

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

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

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

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

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

For Pair MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]) the following chains were created:

We consider the chain MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0]) which results in the following constraint:
(8)    (MULT(x[0], y[0])≥NonInfC∧MULT(x[0], y[0])≥COND_MULT1(>@z(0@z, x[0]), x[0], y[0])∧(U^Increasing(COND_MULT1(>@z(0@z, x[0]), x[0], y[0])), ≥))

We simplified constraint (8) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(9)    ((U^Increasing(COND_MULT1(>@z(0@z, x[0]), 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_MULT1(>@z(0@z, x[0]), 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∧0 ≥ 0∧(U^Increasing(COND_MULT1(>@z(0@z, x[0]), x[0], y[0])), ≥))

We simplified constraint (11) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(12)    (0 = 0∧0 = 0∧0 ≥ 0∧(U^Increasing(COND_MULT1(>@z(0@z, x[0]), x[0], y[0])), ≥)∧0 = 0∧0 ≥ 0∧0 = 0)

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

COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2])
- (x[0] ≥ 0 ⇒ (-1)Bound + x[0] ≥ 0∧1 + (2)x[0] ≥ 0∧(U^Increasing(MULT(-@z(x[2]), y[2])), ≥)∧0 = 0∧0 = 0)
MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0])
- (0 = 0∧0 = 0∧0 ≥ 0∧(U^Increasing(COND_MULT1(>@z(0@z, x[0]), x[0], y[0])), ≥)∧0 = 0∧0 ≥ 0∧0 = 0)

POL(-@z(x₁)) = (-1)x₁
POL(0@z) = 0
POL(MULT(x₁, x₂)) = -1 + (-1)x₁
POL(TRUE) = -1
POL(COND_MULT1(x₁, x₂, x₃)) = -1 + (-1)x₂
POL(FALSE) = -1
POL(undefined) = -1
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2])

The following pairs are in P_bound:

COND_MULT1(TRUE, x[2], y[2]) → MULT(-@z(x[2]), y[2])

The following pairs are in P_≥:

MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0])

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

-@z¹ ↔

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ AND

              ↳ IDP

              ↳ IDP

                ↳ IDependencyGraphProof

                  ↳ IDP

                    ↳ IDPNonInfProof

                      ↳ 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): MULT(x[0], y[0]) → COND_MULT1(>@z(0@z, x[0]), x[0], y[0])

The set Q consists of the following terms:

Cond_mult1(TRUE, x0, x1)
mult(x0, x1)
Cond_mult(TRUE, x0, x1)

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