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_eval(TRUE, x, y) → eval(-@z(x, 1@z), y)
eval(x, y) → Cond_eval(&&(&&(>@z(x, 0@z), !(=@z(x, 0@z))), <=@z(x, y)), x, y)
Cond_eval1(TRUE, x, y) → eval(y, y)
eval(x, y) → Cond_eval1(&&(&&(>@z(x, 0@z), !(=@z(x, 0@z))), >@z(x, y)), x, y)
The set Q consists of the following terms:
Cond_eval(TRUE, x0, x1)
eval(x0, x1)
Cond_eval1(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_eval(TRUE, x, y) → eval(-@z(x, 1@z), y)
eval(x, y) → Cond_eval(&&(&&(>@z(x, 0@z), !(=@z(x, 0@z))), <=@z(x, y)), x, y)
Cond_eval1(TRUE, x, y) → eval(y, y)
eval(x, y) → Cond_eval1(&&(&&(>@z(x, 0@z), !(=@z(x, 0@z))), >@z(x, y)), x, y)
The integer pair graph contains the following rules and edges:
(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])
(1): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])
(2): COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2])
(3): COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3])
(0) -> (3), if ((x[0] →* x[3])∧(y[0] →* y[3])∧(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])) →* TRUE))
(1) -> (2), if ((x[1] →* x[2])∧(y[1] →* y[2])∧(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])) →* TRUE))
(2) -> (0), if ((y[2] →* y[0])∧(-@z(x[2], 1@z) →* x[0]))
(2) -> (1), if ((y[2] →* y[1])∧(-@z(x[2], 1@z) →* x[1]))
(3) -> (0), if ((y[3] →* y[0])∧(y[3] →* x[0]))
(3) -> (1), if ((y[3] →* y[1])∧(y[3] →* x[1]))
The set Q consists of the following terms:
Cond_eval(TRUE, x0, x1)
eval(x0, x1)
Cond_eval1(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): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])
(1): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])
(2): COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2])
(3): COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3])
(0) -> (3), if ((x[0] →* x[3])∧(y[0] →* y[3])∧(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])) →* TRUE))
(1) -> (2), if ((x[1] →* x[2])∧(y[1] →* y[2])∧(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])) →* TRUE))
(2) -> (0), if ((y[2] →* y[0])∧(-@z(x[2], 1@z) →* x[0]))
(2) -> (1), if ((y[2] →* y[1])∧(-@z(x[2], 1@z) →* x[1]))
(3) -> (0), if ((y[3] →* y[0])∧(y[3] →* x[0]))
(3) -> (1), if ((y[3] →* y[1])∧(y[3] →* x[1]))
The set Q consists of the following terms:
Cond_eval(TRUE, x0, x1)
eval(x0, x1)
Cond_eval1(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 EVAL(x, y) → COND_EVAL1(&&(&&(>@z(x, 0@z), !(=@z(x, 0@z))), >@z(x, y)), x, y) the following chains were created:
- We consider the chain EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0]) which results in the following constraint:
(1) (EVAL(x[0], y[0])≥NonInfC∧EVAL(x[0], y[0])≥COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])∧(UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])), ≥))
We simplified constraint (1) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(2) ((UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[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) ((UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[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) ((UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (4) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(5) (0 ≥ 0∧0 = 0∧(UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])), ≥)∧0 = 0∧0 = 0∧0 = 0∧0 ≥ 0)
For Pair EVAL(x, y) → COND_EVAL(&&(&&(>@z(x, 0@z), !(=@z(x, 0@z))), <=@z(x, y)), x, y) the following chains were created:
- We consider the chain EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1]) which results in the following constraint:
(6) (EVAL(x[1], y[1])≥NonInfC∧EVAL(x[1], y[1])≥COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])∧(UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥))
We simplified constraint (6) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(7) ((UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (7) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(8) ((UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (8) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(9) (0 ≥ 0∧(UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥)∧0 ≥ 0)
We simplified constraint (9) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(10) (0 = 0∧0 = 0∧0 = 0∧(UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥)∧0 = 0∧0 ≥ 0∧0 ≥ 0)
For Pair COND_EVAL(TRUE, x, y) → EVAL(-@z(x, 1@z), y) the following chains were created:
- We consider the chain EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1]), COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2]), EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1]) which results in the following constraint:
(11) (&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1]))=TRUE∧x[1]=x[2]∧-@z(x[2], 1@z)=x[1]1∧y[1]=y[2]∧y[2]=y[1]1 ⇒ COND_EVAL(TRUE, x[2], y[2])≥NonInfC∧COND_EVAL(TRUE, x[2], y[2])≥EVAL(-@z(x[2], 1@z), y[2])∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
We simplified constraint (11) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraints:
(12) (<=@z(x[1], y[1])=TRUE∧>@z(x[1], 0@z)=TRUE ⇒ COND_EVAL(TRUE, x[1], y[1])≥NonInfC∧COND_EVAL(TRUE, x[1], y[1])≥EVAL(-@z(x[1], 1@z), y[1])∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
(13) (<=@z(x[1], y[1])=TRUE∧>@z(x[1], 0@z)=TRUE∧<@z(x[1], 0@z)=TRUE ⇒ COND_EVAL(TRUE, x[1], y[1])≥NonInfC∧COND_EVAL(TRUE, x[1], y[1])≥EVAL(-@z(x[1], 1@z), y[1])∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
We simplified constraint (12) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(14) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (13) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(15) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0∧-1 + (-1)x[1] ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (14) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(16) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (15) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(17) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0∧-1 + (-1)x[1] ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (16) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(18) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
We simplified constraint (17) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(19) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0∧-1 + (-1)x[1] ≥ 0 ⇒ 0 ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0)
We simplified constraint (18) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(20) (-1 + y[1] + (-1)x[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
We solved constraint (19) using rule (IDP_SMT_SPLIT).We simplified constraint (20) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(21) (y[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
- We consider the chain EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1]), COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2]), EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0]) which results in the following constraint:
(22) (&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1]))=TRUE∧x[1]=x[2]∧y[2]=y[0]∧y[1]=y[2]∧-@z(x[2], 1@z)=x[0] ⇒ COND_EVAL(TRUE, x[2], y[2])≥NonInfC∧COND_EVAL(TRUE, x[2], y[2])≥EVAL(-@z(x[2], 1@z), y[2])∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
We simplified constraint (22) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraints:
(23) (<=@z(x[1], y[1])=TRUE∧>@z(x[1], 0@z)=TRUE ⇒ COND_EVAL(TRUE, x[1], y[1])≥NonInfC∧COND_EVAL(TRUE, x[1], y[1])≥EVAL(-@z(x[1], 1@z), y[1])∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
(24) (<=@z(x[1], y[1])=TRUE∧>@z(x[1], 0@z)=TRUE∧<@z(x[1], 0@z)=TRUE ⇒ COND_EVAL(TRUE, x[1], y[1])≥NonInfC∧COND_EVAL(TRUE, x[1], y[1])≥EVAL(-@z(x[1], 1@z), y[1])∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
We simplified constraint (23) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(25) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (24) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(26) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0∧-1 + (-1)x[1] ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (25) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(27) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (26) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(28) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0∧-1 + (-1)x[1] ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (27) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(29) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0 ⇒ 0 ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0)
We simplified constraint (28) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(30) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0∧-1 + (-1)x[1] ≥ 0 ⇒ 0 ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0)
We simplified constraint (29) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(31) (-1 + y[1] + (-1)x[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0)
We solved constraint (30) using rule (IDP_SMT_SPLIT).We simplified constraint (31) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(32) (y[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0)
For Pair COND_EVAL1(TRUE, x, y) → EVAL(y, y) the following chains were created:
- We consider the chain EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0]), COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3]), EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0]) which results in the following constraint:
(33) (y[0]=y[3]∧y[3]=x[0]1∧x[0]=x[3]∧&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0]))=TRUE∧y[3]=y[0]1 ⇒ COND_EVAL1(TRUE, x[3], y[3])≥NonInfC∧COND_EVAL1(TRUE, x[3], y[3])≥EVAL(y[3], y[3])∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We simplified constraint (33) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraints:
(34) (>@z(x[0], y[0])=TRUE∧>@z(x[0], 0@z)=TRUE ⇒ COND_EVAL1(TRUE, x[0], y[0])≥NonInfC∧COND_EVAL1(TRUE, x[0], y[0])≥EVAL(y[0], y[0])∧(UIncreasing(EVAL(y[3], y[3])), ≥))
(35) (>@z(x[0], y[0])=TRUE∧>@z(x[0], 0@z)=TRUE∧<@z(x[0], 0@z)=TRUE ⇒ COND_EVAL1(TRUE, x[0], y[0])≥NonInfC∧COND_EVAL1(TRUE, x[0], y[0])≥EVAL(y[0], y[0])∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We simplified constraint (34) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(36) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(-1)y[0] + x[0] ≥ 0)
We simplified constraint (35) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(37) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(-1)y[0] + x[0] ≥ 0)
We simplified constraint (36) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(38) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(-1)y[0] + x[0] ≥ 0)
We simplified constraint (37) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(39) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(-1)y[0] + x[0] ≥ 0)
We simplified constraint (38) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(40) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0 ⇒ (-1)y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0)
We simplified constraint (39) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(41) (-1 + (-1)x[0] ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0 ⇒ 1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(-1)y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We simplified constraint (40) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(42) (x[0] + (-1)y[0] ≥ 0∧x[0] ≥ 0 ⇒ 1 + (-1)y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + (-1)y[0] + x[0] ≥ 0)
We solved constraint (41) using rule (IDP_SMT_SPLIT).We simplified constraint (42) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(43) (x[0] + (-1)y[0] ≥ 0∧x[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + (-1)y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + (-1)y[0] + x[0] ≥ 0)
(44) (x[0] + y[0] ≥ 0∧x[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + y[0] + x[0] ≥ 0)
We simplified constraint (43) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(45) (x[0] ≥ 0∧y[0] + x[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + x[0] ≥ 0)
- We consider the chain EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0]), COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3]), EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1]) which results in the following constraint:
(46) (y[0]=y[3]∧y[3]=y[1]∧x[0]=x[3]∧y[3]=x[1]∧&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0]))=TRUE ⇒ COND_EVAL1(TRUE, x[3], y[3])≥NonInfC∧COND_EVAL1(TRUE, x[3], y[3])≥EVAL(y[3], y[3])∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We simplified constraint (46) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraints:
(47) (>@z(x[0], y[0])=TRUE∧>@z(x[0], 0@z)=TRUE ⇒ COND_EVAL1(TRUE, x[0], y[0])≥NonInfC∧COND_EVAL1(TRUE, x[0], y[0])≥EVAL(y[0], y[0])∧(UIncreasing(EVAL(y[3], y[3])), ≥))
(48) (>@z(x[0], y[0])=TRUE∧>@z(x[0], 0@z)=TRUE∧<@z(x[0], 0@z)=TRUE ⇒ COND_EVAL1(TRUE, x[0], y[0])≥NonInfC∧COND_EVAL1(TRUE, x[0], y[0])≥EVAL(y[0], y[0])∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We simplified constraint (47) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(49) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(-1)y[0] + x[0] ≥ 0)
We simplified constraint (48) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(50) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(-1)y[0] + x[0] ≥ 0)
We simplified constraint (49) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(51) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(-1)y[0] + x[0] ≥ 0)
We simplified constraint (50) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(52) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(-1)y[0] + x[0] ≥ 0)
We simplified constraint (51) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(53) (x[0] + -1 ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0 ⇒ (-1)y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0)
We simplified constraint (52) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(54) (x[0] + -1 ≥ 0∧-1 + (-1)x[0] ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0 ⇒ 1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(-1)y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We simplified constraint (53) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(55) (x[0] ≥ 0∧x[0] + (-1)y[0] ≥ 0 ⇒ 1 + (-1)y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + (-1)y[0] + x[0] ≥ 0)
We solved constraint (54) using rule (IDP_SMT_SPLIT).We simplified constraint (55) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(56) (x[0] ≥ 0∧x[0] + (-1)y[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + (-1)y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + (-1)y[0] + x[0] ≥ 0)
(57) (x[0] ≥ 0∧x[0] + y[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + y[0] + x[0] ≥ 0)
We simplified constraint (56) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(58) (y[0] + x[0] ≥ 0∧x[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + x[0] ≥ 0)
To summarize, we get the following constraints P≥ for the following pairs.
- EVAL(x, y) → COND_EVAL1(&&(&&(>@z(x, 0@z), !(=@z(x, 0@z))), >@z(x, y)), x, y)
- (0 ≥ 0∧0 = 0∧(UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])), ≥)∧0 = 0∧0 = 0∧0 = 0∧0 ≥ 0)
- EVAL(x, y) → COND_EVAL(&&(&&(>@z(x, 0@z), !(=@z(x, 0@z))), <=@z(x, y)), x, y)
- (0 = 0∧0 = 0∧0 = 0∧(UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥)∧0 = 0∧0 ≥ 0∧0 ≥ 0)
- COND_EVAL(TRUE, x, y) → EVAL(-@z(x, 1@z), y)
- (y[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
- (y[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0)
- COND_EVAL1(TRUE, x, y) → EVAL(y, y)
- (x[0] + y[0] ≥ 0∧x[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + y[0] + x[0] ≥ 0)
- (x[0] ≥ 0∧y[0] + x[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + x[0] ≥ 0)
- (x[0] ≥ 0∧x[0] + y[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + y[0] + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + y[0] + x[0] ≥ 0)
- (y[0] + x[0] ≥ 0∧x[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧2 + (-1)Bound + x[0] ≥ 0)
The constraints for P> respective Pbound 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 Pbound.
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(COND_EVAL1(x1, x2, x3)) = (-1)x3 + x2 + (-1)x1
POL(TRUE) = -1
POL(&&(x1, x2)) = -1
POL(COND_EVAL(x1, x2, x3)) = -1 + (-1)x3 + x2 + (-1)x1
POL(!(x1)) = -1
POL(FALSE) = -1
POL(>@z(x1, x2)) = -1
POL(=@z(x1, x2)) = -1
POL(EVAL(x1, x2)) = 1 + (-1)x2 + x1
POL(1@z) = 1
POL(undefined) = -1
The following pairs are in P>:
EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])
The following pairs are in Pbound:
COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3])
The following pairs are in P≥:
EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])
COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2])
COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3])
At least the following rules have been oriented under context sensitive arithmetic replacement:
&&(FALSE, FALSE)1 ↔ FALSE1
-@z1 ↔
&&(TRUE, TRUE)1 ↔ TRUE1
&&(TRUE, FALSE)1 ↔ FALSE1
&&(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): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])
(1): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])
(2): COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2])
(1) -> (2), if ((x[1] →* x[2])∧(y[1] →* y[2])∧(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])) →* TRUE))
(2) -> (0), if ((y[2] →* y[0])∧(-@z(x[2], 1@z) →* x[0]))
(2) -> (1), if ((y[2] →* y[1])∧(-@z(x[2], 1@z) →* x[1]))
The set Q consists of the following terms:
Cond_eval(TRUE, x0, x1)
eval(x0, x1)
Cond_eval1(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): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])
(2): COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2])
(1) -> (2), if ((x[1] →* x[2])∧(y[1] →* y[2])∧(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])) →* TRUE))
(2) -> (1), if ((y[2] →* y[1])∧(-@z(x[2], 1@z) →* x[1]))
The set Q consists of the following terms:
Cond_eval(TRUE, x0, x1)
eval(x0, x1)
Cond_eval1(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 EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1]) the following chains were created:
- We consider the chain EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1]) which results in the following constraint:
(1) (EVAL(x[1], y[1])≥NonInfC∧EVAL(x[1], y[1])≥COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])∧(UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥))
We simplified constraint (1) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(2) ((UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (2) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(3) ((UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥)∧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∧(UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥))
We simplified constraint (4) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(5) (0 = 0∧0 = 0∧0 ≥ 0∧(UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥)∧0 = 0∧0 = 0∧0 ≥ 0)
For Pair COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2]) the following chains were created:
- We consider the chain EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1]), COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2]), EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1]) which results in the following constraint:
(6) (&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1]))=TRUE∧x[1]=x[2]∧-@z(x[2], 1@z)=x[1]1∧y[1]=y[2]∧y[2]=y[1]1 ⇒ COND_EVAL(TRUE, x[2], y[2])≥NonInfC∧COND_EVAL(TRUE, x[2], y[2])≥EVAL(-@z(x[2], 1@z), y[2])∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
We simplified constraint (6) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraints:
(7) (<=@z(x[1], y[1])=TRUE∧>@z(x[1], 0@z)=TRUE ⇒ COND_EVAL(TRUE, x[1], y[1])≥NonInfC∧COND_EVAL(TRUE, x[1], y[1])≥EVAL(-@z(x[1], 1@z), y[1])∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
(8) (<=@z(x[1], y[1])=TRUE∧>@z(x[1], 0@z)=TRUE∧<@z(x[1], 0@z)=TRUE ⇒ COND_EVAL(TRUE, x[1], y[1])≥NonInfC∧COND_EVAL(TRUE, x[1], y[1])≥EVAL(-@z(x[1], 1@z), y[1])∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(9) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧1 + (-1)Bound + y[1] + x[1] ≥ 0∧0 ≥ 0)
We simplified constraint (8) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(10) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0∧-1 + (-1)x[1] ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧1 + (-1)Bound + y[1] + x[1] ≥ 0∧0 ≥ 0)
We simplified constraint (9) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(11) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧1 + (-1)Bound + y[1] + x[1] ≥ 0∧0 ≥ 0)
We simplified constraint (10) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(12) (y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0∧-1 + (-1)x[1] ≥ 0 ⇒ (UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧1 + (-1)Bound + y[1] + x[1] ≥ 0∧0 ≥ 0)
We simplified constraint (11) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(13) (x[1] + -1 ≥ 0∧y[1] + (-1)x[1] ≥ 0 ⇒ 1 + (-1)Bound + y[1] + x[1] ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0)
We simplified constraint (12) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(14) (-1 + (-1)x[1] ≥ 0∧y[1] + (-1)x[1] ≥ 0∧x[1] + -1 ≥ 0 ⇒ 0 ≥ 0∧1 + (-1)Bound + y[1] + x[1] ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥))
We simplified constraint (13) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(15) (x[1] ≥ 0∧-1 + y[1] + (-1)x[1] ≥ 0 ⇒ 2 + (-1)Bound + y[1] + x[1] ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0)
We solved constraint (14) using rule (IDP_SMT_SPLIT).We simplified constraint (15) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(16) (x[1] ≥ 0∧y[1] ≥ 0 ⇒ 3 + (-1)Bound + (2)x[1] + y[1] ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0)
To summarize, we get the following constraints P≥ for the following pairs.
- EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])
- (0 = 0∧0 = 0∧0 ≥ 0∧(UIncreasing(COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])), ≥)∧0 = 0∧0 = 0∧0 ≥ 0)
- COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2])
- (x[1] ≥ 0∧y[1] ≥ 0 ⇒ 3 + (-1)Bound + (2)x[1] + y[1] ≥ 0∧(UIncreasing(EVAL(-@z(x[2], 1@z), y[2])), ≥)∧0 ≥ 0)
The constraints for P> respective Pbound 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 Pbound.
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) = 2
POL(&&(x1, x2)) = 0
POL(COND_EVAL(x1, x2, x3)) = 1 + x3 + x2
POL(!(x1)) = -1
POL(FALSE) = -1
POL(>@z(x1, x2)) = -1
POL(=@z(x1, x2)) = -1
POL(EVAL(x1, x2)) = 1 + x2 + x1
POL(1@z) = 1
POL(undefined) = -1
The following pairs are in P>:
COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2])
The following pairs are in Pbound:
COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2])
The following pairs are in P≥:
EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])
At least the following rules have been oriented under context sensitive arithmetic replacement:
&&(FALSE, FALSE)1 → FALSE1
-@z1 ↔
&&(FALSE, TRUE)1 → FALSE1
↳ 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:
(1): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(x[1], 0@z), !(=@z(x[1], 0@z))), <=@z(x[1], y[1])), x[1], y[1])
The set Q consists of the following terms:
Cond_eval(TRUE, x0, x1)
eval(x0, x1)
Cond_eval1(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): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])
(2): COND_EVAL(TRUE, x[2], y[2]) → EVAL(-@z(x[2], 1@z), y[2])
(3): COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3])
(0) -> (3), if ((x[0] →* x[3])∧(y[0] →* y[3])∧(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])) →* TRUE))
(2) -> (0), if ((y[2] →* y[0])∧(-@z(x[2], 1@z) →* x[0]))
(3) -> (0), if ((y[3] →* y[0])∧(y[3] →* x[0]))
The set Q consists of the following terms:
Cond_eval(TRUE, x0, x1)
eval(x0, x1)
Cond_eval1(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:
(3): COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3])
(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])
(0) -> (3), if ((x[0] →* x[3])∧(y[0] →* y[3])∧(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])) →* TRUE))
(3) -> (0), if ((y[3] →* y[0])∧(y[3] →* x[0]))
The set Q consists of the following terms:
Cond_eval(TRUE, x0, x1)
eval(x0, x1)
Cond_eval1(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_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3]) the following chains were created:
- We consider the chain EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0]), COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3]), EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0]) which results in the following constraint:
(1) (y[0]=y[3]∧y[3]=x[0]1∧x[0]=x[3]∧&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0]))=TRUE∧y[3]=y[0]1 ⇒ COND_EVAL1(TRUE, x[3], y[3])≥NonInfC∧COND_EVAL1(TRUE, x[3], y[3])≥EVAL(y[3], y[3])∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraints:
(2) (>@z(x[0], y[0])=TRUE∧>@z(x[0], 0@z)=TRUE ⇒ COND_EVAL1(TRUE, x[0], y[0])≥NonInfC∧COND_EVAL1(TRUE, x[0], y[0])≥EVAL(y[0], y[0])∧(UIncreasing(EVAL(y[3], y[3])), ≥))
(3) (>@z(x[0], y[0])=TRUE∧>@z(x[0], 0@z)=TRUE∧<@z(x[0], 0@z)=TRUE ⇒ COND_EVAL1(TRUE, x[0], y[0])≥NonInfC∧COND_EVAL1(TRUE, x[0], y[0])≥EVAL(y[0], y[0])∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(4) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧-1 + (-1)Bound + (2)x[0] ≥ 0∧-1 + (-2)y[0] + (2)x[0] ≥ 0)
We simplified constraint (3) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(5) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧-1 + (-1)Bound + (2)x[0] ≥ 0∧-1 + (-2)y[0] + (2)x[0] ≥ 0)
We simplified constraint (4) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(6) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧-1 + (-1)Bound + (2)x[0] ≥ 0∧-1 + (-2)y[0] + (2)x[0] ≥ 0)
We simplified constraint (5) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(7) (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + -1 ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ (UIncreasing(EVAL(y[3], y[3])), ≥)∧-1 + (-1)Bound + (2)x[0] ≥ 0∧-1 + (-2)y[0] + (2)x[0] ≥ 0)
We simplified constraint (6) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(8) (x[0] + -1 ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0 ⇒ -1 + (-2)y[0] + (2)x[0] ≥ 0∧-1 + (-1)Bound + (2)x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We simplified constraint (7) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(9) (x[0] + -1 ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0∧-1 + (-1)x[0] ≥ 0 ⇒ -1 + (-2)y[0] + (2)x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥)∧-1 + (-1)Bound + (2)x[0] ≥ 0)
We simplified constraint (8) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(10) (x[0] ≥ 0∧x[0] + (-1)y[0] ≥ 0 ⇒ 1 + (-2)y[0] + (2)x[0] ≥ 0∧1 + (-1)Bound + (2)x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We solved constraint (9) using rule (IDP_SMT_SPLIT).We simplified constraint (10) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(11) (x[0] ≥ 0∧x[0] + (-1)y[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + (-2)y[0] + (2)x[0] ≥ 0∧1 + (-1)Bound + (2)x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥))
(12) (x[0] ≥ 0∧x[0] + y[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + (2)y[0] + (2)x[0] ≥ 0∧1 + (-1)Bound + (2)x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥))
We simplified constraint (11) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(13) (y[0] + x[0] ≥ 0∧x[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + (2)x[0] ≥ 0∧1 + (-1)Bound + (2)y[0] + (2)x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥))
For Pair EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0]) the following chains were created:
- We consider the chain EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0]) which results in the following constraint:
(14) (EVAL(x[0], y[0])≥NonInfC∧EVAL(x[0], y[0])≥COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])∧(UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])), ≥))
We simplified constraint (14) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(15) ((UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (15) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(16) ((UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (16) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(17) (0 ≥ 0∧(UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])), ≥)∧0 ≥ 0)
We simplified constraint (17) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(18) (0 ≥ 0∧0 ≥ 0∧(UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), 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_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3])
- (x[0] ≥ 0∧x[0] + y[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + (2)y[0] + (2)x[0] ≥ 0∧1 + (-1)Bound + (2)x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥))
- (y[0] + x[0] ≥ 0∧x[0] ≥ 0∧y[0] ≥ 0 ⇒ 1 + (2)x[0] ≥ 0∧1 + (-1)Bound + (2)y[0] + (2)x[0] ≥ 0∧(UIncreasing(EVAL(y[3], y[3])), ≥))
- EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])
- (0 ≥ 0∧0 ≥ 0∧(UIncreasing(COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])), ≥)∧0 = 0∧0 = 0∧0 = 0∧0 = 0)
The constraints for P> respective Pbound 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 Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:
POL(=@z(x1, x2)) = -1
POL(0@z) = 0
POL(COND_EVAL1(x1, x2, x3)) = -1 + (2)x2
POL(TRUE) = 2
POL(&&(x1, x2)) = -1
POL(EVAL(x1, x2)) = -1 + (2)x1
POL(!(x1)) = -1
POL(FALSE) = -1
POL(undefined) = -1
POL(>@z(x1, x2)) = -1
The following pairs are in P>:
COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3])
The following pairs are in Pbound:
COND_EVAL1(TRUE, x[3], y[3]) → EVAL(y[3], y[3])
The following pairs are in P≥:
EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])
At least the following rules have been oriented under context sensitive arithmetic replacement:
&&(FALSE, FALSE)1 → FALSE1
&&(FALSE, TRUE)1 → FALSE1
&&(TRUE, FALSE)1 ↔ FALSE1
↳ 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): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(x[0], 0@z), !(=@z(x[0], 0@z))), >@z(x[0], y[0])), x[0], y[0])
The set Q consists of the following terms:
Cond_eval(TRUE, x0, x1)
eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.