### (0) Obligation:

JBC Problem based on JBC Program:
`No human-readable program information known.`

Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: example_1/Test

### (1) JBC2FIG (SOUND transformation)

Constructed FIGraph.

### (2) Obligation:

FIGraph based on JBC Program:
Graph of 165 nodes with 3 SCCs.

### (3) FIGtoITRSProof (SOUND transformation)

Transformed FIGraph to ITRS rules

### (5) Obligation:

ITRS problem:

The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The TRS R consists of the following rules:
Load850(1000, 1000, i47, i48) → Cond_Load850(i48 >= 0 && i48 <= 1000 && i48 + 3 > 0, 1000, 1000, i47, i48)
Cond_Load850(TRUE, 1000, 1000, i47, i48) → Load850(1000, 1000, i47 + i48, i48 + 3)
The set Q consists of the following terms:

### (6) GroundTermsRemoverProof (EQUIVALENT transformation)

Some arguments are removed because they always contain the same ground term.
We removed the following ground terms:
• 1000

We removed arguments according to the following replacements:

### (7) Obligation:

ITRS problem:

The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The TRS R consists of the following rules:
Load850(i47, i48) → Cond_Load850(i48 >= 0 && i48 <= 1000 && i48 + 3 > 0, i47, i48)
The set Q consists of the following terms:

### (9) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Boolean, Integer

The ITRS R consists of the following rules:
Load850(i47, i48) → Cond_Load850(i48 >= 0 && i48 <= 1000 && i48 + 3 > 0, i47, i48)

The integer pair graph contains the following rules and edges:
(0): LOAD850(i47[0], i48[0]) → COND_LOAD850(i48[0] >= 0 && i48[0] <= 1000 && i48[0] + 3 > 0, i47[0], i48[0])

(0) -> (1), if ((i48[0]* i48[1])∧(i47[0]* i47[1])∧(i48[0] >= 0 && i48[0] <= 1000 && i48[0] + 3 > 0* TRUE))

(1) -> (0), if ((i48[1] + 3* i48[0])∧(i47[1] + i48[1]* i47[0]))

The set Q consists of the following terms:

### (10) UsableRulesProof (EQUIVALENT transformation)

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.

### (11) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Boolean, Integer

R is empty.

The integer pair graph contains the following rules and edges:
(0): LOAD850(i47[0], i48[0]) → COND_LOAD850(i48[0] >= 0 && i48[0] <= 1000 && i48[0] + 3 > 0, i47[0], i48[0])

(0) -> (1), if ((i48[0]* i48[1])∧(i47[0]* i47[1])∧(i48[0] >= 0 && i48[0] <= 1000 && i48[0] + 3 > 0* TRUE))

(1) -> (0), if ((i48[1] + 3* i48[0])∧(i47[1] + i48[1]* i47[0]))

The set Q consists of the following terms:

### (12) IDPNonInfProof (SOUND transformation)

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 LOAD850(i47, i48) → COND_LOAD850(&&(&&(>=(i48, 0), <=(i48, 1000)), >(+(i48, 3), 0)), i47, i48) the following chains were created:
• We consider the chain LOAD850(i47[0], i48[0]) → COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0]), COND_LOAD850(TRUE, i47[1], i48[1]) → LOAD850(+(i47[1], i48[1]), +(i48[1], 3)) which results in the following constraint:

(1)    (i48[0]=i48[1]i47[0]=i47[1]&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0))=TRUELOAD850(i47[0], i48[0])≥NonInfC∧LOAD850(i47[0], i48[0])≥COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])∧(UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥))

We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint:

(2)    (>(+(i48[0], 3), 0)=TRUE>=(i48[0], 0)=TRUE<=(i48[0], 1000)=TRUELOAD850(i47[0], i48[0])≥NonInfC∧LOAD850(i47[0], i48[0])≥COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])∧(UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥))

We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:

(3)    (i48[0] + [2] ≥ 0∧i48[0] ≥ 0∧[1000] + [-1]i48[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i48[0] ≥ 0∧[(-1)bso_12] ≥ 0)

We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:

(4)    (i48[0] + [2] ≥ 0∧i48[0] ≥ 0∧[1000] + [-1]i48[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i48[0] ≥ 0∧[(-1)bso_12] ≥ 0)

We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:

(5)    (i48[0] + [2] ≥ 0∧i48[0] ≥ 0∧[1000] + [-1]i48[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i48[0] ≥ 0∧[(-1)bso_12] ≥ 0)

We simplified constraint (5) using rules (IDP_UNRESTRICTED_VARS), (IDP_POLY_GCD) which results in the following new constraint:

(6)    (i48[0] ≥ 0∧[1000] + [-1]i48[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥)∧0 = 0∧[(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i48[0] ≥ 0∧0 = 0∧[(-1)bso_12] ≥ 0)

For Pair COND_LOAD850(TRUE, i47, i48) → LOAD850(+(i47, i48), +(i48, 3)) the following chains were created:
• We consider the chain COND_LOAD850(TRUE, i47[1], i48[1]) → LOAD850(+(i47[1], i48[1]), +(i48[1], 3)) which results in the following constraint:

We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:

(8)    ((UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥)∧[3 + (-1)bso_14] ≥ 0)

We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:

(9)    ((UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥)∧[3 + (-1)bso_14] ≥ 0)

We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:

(10)    ((UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥)∧[3 + (-1)bso_14] ≥ 0)

We simplified constraint (10) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:

(11)    ((UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥)∧0 = 0∧0 = 0∧[3 + (-1)bso_14] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD850(i47, i48) → COND_LOAD850(&&(&&(>=(i48, 0), <=(i48, 1000)), >(+(i48, 3), 0)), i47, i48)
• (i48[0] ≥ 0∧[1000] + [-1]i48[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥)∧0 = 0∧[(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i48[0] ≥ 0∧0 = 0∧[(-1)bso_12] ≥ 0)

• ((UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥)∧0 = 0∧0 = 0∧[3 + (-1)bso_14] ≥ 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(TRUE) = 0
POL(FALSE) = 0
POL(LOAD850(x1, x2)) = [-1] + [-1]x2
POL(COND_LOAD850(x1, x2, x3)) = [-1] + [-1]x3
POL(&&(x1, x2)) = [-1]
POL(>=(x1, x2)) = [-1]
POL(0) = 0
POL(<=(x1, x2)) = [-1]
POL(1000) = [1000]
POL(>(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(3) = [3]

The following pairs are in P>:

The following pairs are in Pbound:

LOAD850(i47[0], i48[0]) → COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])

The following pairs are in P:

LOAD850(i47[0], i48[0]) → COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])

There are no usable rules.

### (14) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Boolean, Integer

R is empty.

The integer pair graph contains the following rules and edges:
(0): LOAD850(i47[0], i48[0]) → COND_LOAD850(i48[0] >= 0 && i48[0] <= 1000 && i48[0] + 3 > 0, i47[0], i48[0])

The set Q consists of the following terms:

### (15) IDependencyGraphProof (EQUIVALENT transformation)

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

### (17) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Integer

R is empty.

The integer pair graph contains the following rules and edges:

The set Q consists of the following terms:

### (18) IDependencyGraphProof (EQUIVALENT transformation)

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

### (20) Obligation:

ITRS problem:

The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The TRS R consists of the following rules:
Load553(1000, 1000, 1000, i20, i21) → Cond_Load553(i21 >= 0 && i21 <= 1000 && i21 + 2 > 0, 1000, 1000, 1000, i20, i21)
Cond_Load553(TRUE, 1000, 1000, 1000, i20, i21) → Load553(1000, 1000, 1000, i20 + i21, i21 + 2)
The set Q consists of the following terms:
Cond_Load553(TRUE, 1000, 1000, 1000, x0, x1)

### (21) GroundTermsRemoverProof (EQUIVALENT transformation)

Some arguments are removed because they always contain the same ground term.
We removed the following ground terms:
• 1000

We removed arguments according to the following replacements:

### (22) Obligation:

ITRS problem:

The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The TRS R consists of the following rules:
Load553(i20, i21) → Cond_Load553(i21 >= 0 && i21 <= 1000 && i21 + 2 > 0, i20, i21)
The set Q consists of the following terms:

### (24) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Boolean, Integer

The ITRS R consists of the following rules:
Load553(i20, i21) → Cond_Load553(i21 >= 0 && i21 <= 1000 && i21 + 2 > 0, i20, i21)

The integer pair graph contains the following rules and edges:
(0): LOAD553(i20[0], i21[0]) → COND_LOAD553(i21[0] >= 0 && i21[0] <= 1000 && i21[0] + 2 > 0, i20[0], i21[0])

(0) -> (1), if ((i21[0]* i21[1])∧(i20[0]* i20[1])∧(i21[0] >= 0 && i21[0] <= 1000 && i21[0] + 2 > 0* TRUE))

(1) -> (0), if ((i20[1] + i21[1]* i20[0])∧(i21[1] + 2* i21[0]))

The set Q consists of the following terms:

### (25) UsableRulesProof (EQUIVALENT transformation)

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.

### (26) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Boolean, Integer

R is empty.

The integer pair graph contains the following rules and edges:
(0): LOAD553(i20[0], i21[0]) → COND_LOAD553(i21[0] >= 0 && i21[0] <= 1000 && i21[0] + 2 > 0, i20[0], i21[0])

(0) -> (1), if ((i21[0]* i21[1])∧(i20[0]* i20[1])∧(i21[0] >= 0 && i21[0] <= 1000 && i21[0] + 2 > 0* TRUE))

(1) -> (0), if ((i20[1] + i21[1]* i20[0])∧(i21[1] + 2* i21[0]))

The set Q consists of the following terms:

### (27) IDPNonInfProof (SOUND transformation)

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 LOAD553(i20, i21) → COND_LOAD553(&&(&&(>=(i21, 0), <=(i21, 1000)), >(+(i21, 2), 0)), i20, i21) the following chains were created:
• We consider the chain LOAD553(i20[0], i21[0]) → COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0]), COND_LOAD553(TRUE, i20[1], i21[1]) → LOAD553(+(i20[1], i21[1]), +(i21[1], 2)) which results in the following constraint:

(1)    (i21[0]=i21[1]i20[0]=i20[1]&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0))=TRUELOAD553(i20[0], i21[0])≥NonInfC∧LOAD553(i20[0], i21[0])≥COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])∧(UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥))

We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint:

(2)    (>(+(i21[0], 2), 0)=TRUE>=(i21[0], 0)=TRUE<=(i21[0], 1000)=TRUELOAD553(i20[0], i21[0])≥NonInfC∧LOAD553(i20[0], i21[0])≥COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])∧(UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥))

We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:

(3)    (i21[0] + [1] ≥ 0∧i21[0] ≥ 0∧[1000] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i21[0] ≥ 0∧[1 + (-1)bso_12] ≥ 0)

We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:

(4)    (i21[0] + [1] ≥ 0∧i21[0] ≥ 0∧[1000] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i21[0] ≥ 0∧[1 + (-1)bso_12] ≥ 0)

We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:

(5)    (i21[0] + [1] ≥ 0∧i21[0] ≥ 0∧[1000] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i21[0] ≥ 0∧[1 + (-1)bso_12] ≥ 0)

We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:

(6)    (i21[0] + [1] ≥ 0∧i21[0] ≥ 0∧[1000] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥)∧0 = 0∧[(-1)Bound*bni_11] + [(-1)bni_11]i21[0] ≥ 0∧0 = 0∧[1 + (-1)bso_12] ≥ 0)

For Pair COND_LOAD553(TRUE, i20, i21) → LOAD553(+(i20, i21), +(i21, 2)) the following chains were created:
• We consider the chain COND_LOAD553(TRUE, i20[1], i21[1]) → LOAD553(+(i20[1], i21[1]), +(i21[1], 2)) which results in the following constraint:

We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:

(8)    ((UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥)∧[1 + (-1)bso_14] ≥ 0)

We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:

(9)    ((UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥)∧[1 + (-1)bso_14] ≥ 0)

We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:

(10)    ((UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥)∧[1 + (-1)bso_14] ≥ 0)

We simplified constraint (10) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:

(11)    ((UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_14] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD553(i20, i21) → COND_LOAD553(&&(&&(>=(i21, 0), <=(i21, 1000)), >(+(i21, 2), 0)), i20, i21)
• (i21[0] + [1] ≥ 0∧i21[0] ≥ 0∧[1000] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥)∧0 = 0∧[(-1)Bound*bni_11] + [(-1)bni_11]i21[0] ≥ 0∧0 = 0∧[1 + (-1)bso_12] ≥ 0)

• ((UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_14] ≥ 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(TRUE) = 0
POL(FALSE) = 0
POL(COND_LOAD553(x1, x2, x3)) = [-1] + [-1]x3
POL(&&(x1, x2)) = [-1]
POL(>=(x1, x2)) = [-1]
POL(0) = 0
POL(<=(x1, x2)) = [-1]
POL(1000) = [1000]
POL(>(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(2) = [2]

The following pairs are in P>:

LOAD553(i20[0], i21[0]) → COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])

The following pairs are in Pbound:

LOAD553(i20[0], i21[0]) → COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])

The following pairs are in P:
none

There are no usable rules.

### (29) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:
none

R is empty.

The integer pair graph is empty.

The set Q consists of the following terms:

### (30) IDependencyGraphProof (EQUIVALENT transformation)

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

### (32) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Integer

R is empty.

The integer pair graph contains the following rules and edges:

The set Q consists of the following terms:

### (33) IDependencyGraphProof (EQUIVALENT transformation)

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

### (35) Obligation:

ITRS problem:

The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The TRS R consists of the following rules:
Load257(1000, 1000, 1000, i5, i6) → Cond_Load257(i6 >= 0 && i6 <= 1000 && i6 + 1 > 0, 1000, 1000, 1000, i5, i6)
Cond_Load257(TRUE, 1000, 1000, 1000, i5, i6) → Load257(1000, 1000, 1000, i5 + i6, i6 + 1)
The set Q consists of the following terms:
Cond_Load257(TRUE, 1000, 1000, 1000, x0, x1)

### (36) GroundTermsRemoverProof (EQUIVALENT transformation)

Some arguments are removed because they always contain the same ground term.
We removed the following ground terms:
• 1000

We removed arguments according to the following replacements:

### (37) Obligation:

ITRS problem:

The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The TRS R consists of the following rules:
Load257(i5, i6) → Cond_Load257(i6 >= 0 && i6 <= 1000 && i6 + 1 > 0, i5, i6)
The set Q consists of the following terms:

### (39) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Boolean, Integer

The ITRS R consists of the following rules:
Load257(i5, i6) → Cond_Load257(i6 >= 0 && i6 <= 1000 && i6 + 1 > 0, i5, i6)

The integer pair graph contains the following rules and edges:
(0): LOAD257(i5[0], i6[0]) → COND_LOAD257(i6[0] >= 0 && i6[0] <= 1000 && i6[0] + 1 > 0, i5[0], i6[0])

(0) -> (1), if ((i6[0] >= 0 && i6[0] <= 1000 && i6[0] + 1 > 0* TRUE)∧(i6[0]* i6[1])∧(i5[0]* i5[1]))

(1) -> (0), if ((i5[1] + i6[1]* i5[0])∧(i6[1] + 1* i6[0]))

The set Q consists of the following terms:

### (40) UsableRulesProof (EQUIVALENT transformation)

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.

### (41) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Boolean, Integer

R is empty.

The integer pair graph contains the following rules and edges:
(0): LOAD257(i5[0], i6[0]) → COND_LOAD257(i6[0] >= 0 && i6[0] <= 1000 && i6[0] + 1 > 0, i5[0], i6[0])

(0) -> (1), if ((i6[0] >= 0 && i6[0] <= 1000 && i6[0] + 1 > 0* TRUE)∧(i6[0]* i6[1])∧(i5[0]* i5[1]))

(1) -> (0), if ((i5[1] + i6[1]* i5[0])∧(i6[1] + 1* i6[0]))

The set Q consists of the following terms:

### (42) IDPNonInfProof (SOUND transformation)

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 LOAD257(i5, i6) → COND_LOAD257(&&(&&(>=(i6, 0), <=(i6, 1000)), >(+(i6, 1), 0)), i5, i6) the following chains were created:
• We consider the chain LOAD257(i5[0], i6[0]) → COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0]), COND_LOAD257(TRUE, i5[1], i6[1]) → LOAD257(+(i5[1], i6[1]), +(i6[1], 1)) which results in the following constraint:

(1)    (&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0))=TRUEi6[0]=i6[1]i5[0]=i5[1]LOAD257(i5[0], i6[0])≥NonInfC∧LOAD257(i5[0], i6[0])≥COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])∧(UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥))

We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint:

(2)    (>(+(i6[0], 1), 0)=TRUE>=(i6[0], 0)=TRUE<=(i6[0], 1000)=TRUELOAD257(i5[0], i6[0])≥NonInfC∧LOAD257(i5[0], i6[0])≥COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])∧(UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥))

We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:

(3)    (i6[0] ≥ 0∧i6[0] ≥ 0∧[1000] + [-1]i6[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i6[0] ≥ 0∧[(-1)bso_12] ≥ 0)

We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:

(4)    (i6[0] ≥ 0∧i6[0] ≥ 0∧[1000] + [-1]i6[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i6[0] ≥ 0∧[(-1)bso_12] ≥ 0)

We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:

(5)    (i6[0] ≥ 0∧i6[0] ≥ 0∧[1000] + [-1]i6[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i6[0] ≥ 0∧[(-1)bso_12] ≥ 0)

We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:

(6)    (i6[0] ≥ 0∧i6[0] ≥ 0∧[1000] + [-1]i6[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥)∧0 = 0∧[(-1)Bound*bni_11] + [(-1)bni_11]i6[0] ≥ 0∧0 = 0∧[(-1)bso_12] ≥ 0)

For Pair COND_LOAD257(TRUE, i5, i6) → LOAD257(+(i5, i6), +(i6, 1)) the following chains were created:
• We consider the chain COND_LOAD257(TRUE, i5[1], i6[1]) → LOAD257(+(i5[1], i6[1]), +(i6[1], 1)) which results in the following constraint:

We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:

(8)    ((UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥)∧[1 + (-1)bso_14] ≥ 0)

We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:

(9)    ((UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥)∧[1 + (-1)bso_14] ≥ 0)

We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:

(10)    ((UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥)∧[1 + (-1)bso_14] ≥ 0)

We simplified constraint (10) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:

(11)    ((UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_14] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD257(i5, i6) → COND_LOAD257(&&(&&(>=(i6, 0), <=(i6, 1000)), >(+(i6, 1), 0)), i5, i6)
• (i6[0] ≥ 0∧i6[0] ≥ 0∧[1000] + [-1]i6[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥)∧0 = 0∧[(-1)Bound*bni_11] + [(-1)bni_11]i6[0] ≥ 0∧0 = 0∧[(-1)bso_12] ≥ 0)

• ((UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_14] ≥ 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(TRUE) = 0
POL(FALSE) = 0
POL(&&(x1, x2)) = [-1]
POL(>=(x1, x2)) = [-1]
POL(0) = 0
POL(<=(x1, x2)) = [-1]
POL(1000) = [1000]
POL(>(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(1) = [1]

The following pairs are in P>:

The following pairs are in Pbound:

LOAD257(i5[0], i6[0]) → COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])

The following pairs are in P:

LOAD257(i5[0], i6[0]) → COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])

There are no usable rules.

### (44) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Boolean, Integer

R is empty.

The integer pair graph contains the following rules and edges:
(0): LOAD257(i5[0], i6[0]) → COND_LOAD257(i6[0] >= 0 && i6[0] <= 1000 && i6[0] + 1 > 0, i5[0], i6[0])

The set Q consists of the following terms:

### (45) IDependencyGraphProof (EQUIVALENT transformation)

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

### (47) Obligation:

IDP problem:
The following function symbols are pre-defined:
 != ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean

The following domains are used:

Integer

R is empty.

The integer pair graph contains the following rules and edges: