### (0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: example_1/Test
`package example_1;public class A {	int incr(int i) {		return i=i+1;	}}package example_1;public class B extends A {	int incr(int i) {		return i = i+2;	}}package example_1;public class C extends B {	int incr(int i) {		return i=i+3;	}}package example_1;public class Test {	public int add(int n,A o){		int res=0;		int i=0;		while (i<=n){			res=res+i;			i=o.incr(i);		}    		return res;	}	public static void main(String[] args) {		int test = 1000;		Test testClass = new Test();		A a = new A();		int result1 = testClass.add(test,a);		a = new B();		int result2 = testClass.add(test,a);		a = new C(); 		int result3 = testClass.add(test,a);     		// System.out.println("Result: "+result1 + result2 + result3);	}}`

### (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:
Load1041(1000, 1000, i46, i47) → Cond_Load1041(i47 >= 0 && i47 <= 1000 && i47 + 3 > 0, 1000, 1000, i46, i47)
Cond_Load1041(TRUE, 1000, 1000, i46, i47) → Load1041(1000, 1000, i46 + i47, i47 + 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:
Load1041(i46, i47) → Cond_Load1041(i47 >= 0 && i47 <= 1000 && i47 + 3 > 0, i46, i47)
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:
Load1041(i46, i47) → Cond_Load1041(i47 >= 0 && i47 <= 1000 && i47 + 3 > 0, i46, i47)

The integer pair graph contains the following rules and edges:
(0): LOAD1041(i46[0], i47[0]) → COND_LOAD1041(i47[0] >= 0 && i47[0] <= 1000 && i47[0] + 3 > 0, i46[0], i47[0])

(0) -> (1), if ((i47[0]* i47[1])∧(i46[0]* i46[1])∧(i47[0] >= 0 && i47[0] <= 1000 && i47[0] + 3 > 0* TRUE))

(1) -> (0), if ((i47[1] + 3* i47[0])∧(i46[1] + i47[1]* i46[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): LOAD1041(i46[0], i47[0]) → COND_LOAD1041(i47[0] >= 0 && i47[0] <= 1000 && i47[0] + 3 > 0, i46[0], i47[0])

(0) -> (1), if ((i47[0]* i47[1])∧(i46[0]* i46[1])∧(i47[0] >= 0 && i47[0] <= 1000 && i47[0] + 3 > 0* TRUE))

(1) -> (0), if ((i47[1] + 3* i47[0])∧(i46[1] + i47[1]* i46[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 LOAD1041(i46, i47) → COND_LOAD1041(&&(&&(>=(i47, 0), <=(i47, 1000)), >(+(i47, 3), 0)), i46, i47) the following chains were created:
• We consider the chain LOAD1041(i46[0], i47[0]) → COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0]), COND_LOAD1041(TRUE, i46[1], i47[1]) → LOAD1041(+(i46[1], i47[1]), +(i47[1], 3)) which results in the following constraint:

(1)    (i47[0]=i47[1]i46[0]=i46[1]&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0))=TRUELOAD1041(i46[0], i47[0])≥NonInfC∧LOAD1041(i46[0], i47[0])≥COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])∧(UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥))

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

(2)    (>(+(i47[0], 3), 0)=TRUE>=(i47[0], 0)=TRUE<=(i47[0], 1000)=TRUELOAD1041(i46[0], i47[0])≥NonInfC∧LOAD1041(i46[0], i47[0])≥COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])∧(UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥))

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

(3)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i47[0] ≥ 0∧[(-1)bso_15] ≥ 0)

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

(4)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i47[0] ≥ 0∧[(-1)bso_15] ≥ 0)

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

(5)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i47[0] ≥ 0∧[(-1)bso_15] ≥ 0)

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

(6)    (i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥)∧0 = 0∧[(-1)Bound*bni_14] + [(-1)bni_14]i47[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)

For Pair COND_LOAD1041(TRUE, i46, i47) → LOAD1041(+(i46, i47), +(i47, 3)) the following chains were created:
• We consider the chain LOAD1041(i46[0], i47[0]) → COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0]), COND_LOAD1041(TRUE, i46[1], i47[1]) → LOAD1041(+(i46[1], i47[1]), +(i47[1], 3)), LOAD1041(i46[0], i47[0]) → COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0]) which results in the following constraint:

We simplified constraint (7) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:

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

(9)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i47[0] ≥ 0∧[3 + (-1)bso_17] ≥ 0)

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

(10)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i47[0] ≥ 0∧[3 + (-1)bso_17] ≥ 0)

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

(11)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i47[0] ≥ 0∧[3 + (-1)bso_17] ≥ 0)

We simplified constraint (11) using rules (IDP_UNRESTRICTED_VARS), (IDP_POLY_GCD) which results in the following new constraint:

(12)    (i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥)∧0 = 0∧[(-1)Bound*bni_16] + [(-1)bni_16]i47[0] ≥ 0∧0 = 0∧[3 + (-1)bso_17] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD1041(i46, i47) → COND_LOAD1041(&&(&&(>=(i47, 0), <=(i47, 1000)), >(+(i47, 3), 0)), i46, i47)
• (i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥)∧0 = 0∧[(-1)Bound*bni_14] + [(-1)bni_14]i47[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)

• (i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥)∧0 = 0∧[(-1)Bound*bni_16] + [(-1)bni_16]i47[0] ≥ 0∧0 = 0∧[3 + (-1)bso_17] ≥ 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) = [1]
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:

LOAD1041(i46[0], i47[0]) → COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])

The following pairs are in P:

LOAD1041(i46[0], i47[0]) → COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])

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

FALSE1&&(FALSE, FALSE)1

### (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): LOAD1041(i46[0], i47[0]) → COND_LOAD1041(i47[0] >= 0 && i47[0] <= 1000 && i47[0] + 3 > 0, i46[0], i47[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:
none

R is empty.

The integer pair graph is empty.

The set Q consists of the following terms:

### (18) IDependencyGraphProof (EQUIVALENT transformation)

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

### (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:
Load731(1000, 1000, 1000, i19, i20) → Cond_Load731(i20 >= 0 && i20 <= 1000 && i20 + 2 > 0, 1000, 1000, 1000, i19, i20)
Cond_Load731(TRUE, 1000, 1000, 1000, i19, i20) → Load731(1000, 1000, 1000, i19 + i20, i20 + 2)
The set Q consists of the following terms:
Cond_Load731(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:
Load731(i19, i20) → Cond_Load731(i20 >= 0 && i20 <= 1000 && i20 + 2 > 0, i19, i20)
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:
Load731(i19, i20) → Cond_Load731(i20 >= 0 && i20 <= 1000 && i20 + 2 > 0, i19, i20)

The integer pair graph contains the following rules and edges:
(0): LOAD731(i19[0], i20[0]) → COND_LOAD731(i20[0] >= 0 && i20[0] <= 1000 && i20[0] + 2 > 0, i19[0], i20[0])

(0) -> (1), if ((i19[0]* i19[1])∧(i20[0]* i20[1])∧(i20[0] >= 0 && i20[0] <= 1000 && i20[0] + 2 > 0* TRUE))

(1) -> (0), if ((i19[1] + i20[1]* i19[0])∧(i20[1] + 2* i20[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): LOAD731(i19[0], i20[0]) → COND_LOAD731(i20[0] >= 0 && i20[0] <= 1000 && i20[0] + 2 > 0, i19[0], i20[0])

(0) -> (1), if ((i19[0]* i19[1])∧(i20[0]* i20[1])∧(i20[0] >= 0 && i20[0] <= 1000 && i20[0] + 2 > 0* TRUE))

(1) -> (0), if ((i19[1] + i20[1]* i19[0])∧(i20[1] + 2* i20[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 LOAD731(i19, i20) → COND_LOAD731(&&(&&(>=(i20, 0), <=(i20, 1000)), >(+(i20, 2), 0)), i19, i20) the following chains were created:
• We consider the chain LOAD731(i19[0], i20[0]) → COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0]), COND_LOAD731(TRUE, i19[1], i20[1]) → LOAD731(+(i19[1], i20[1]), +(i20[1], 2)) which results in the following constraint:

(1)    (i19[0]=i19[1]i20[0]=i20[1]&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0))=TRUELOAD731(i19[0], i20[0])≥NonInfC∧LOAD731(i19[0], i20[0])≥COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])∧(UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥))

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

(2)    (>(+(i20[0], 2), 0)=TRUE>=(i20[0], 0)=TRUE<=(i20[0], 1000)=TRUELOAD731(i19[0], i20[0])≥NonInfC∧LOAD731(i19[0], i20[0])≥COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])∧(UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥))

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

(3)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥)∧[(-1)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i20[0] ≥ 0∧[(-1)bso_15] ≥ 0)

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

(4)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥)∧[(-1)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i20[0] ≥ 0∧[(-1)bso_15] ≥ 0)

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

(5)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥)∧[(-1)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i20[0] ≥ 0∧[(-1)bso_15] ≥ 0)

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

(6)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥)∧0 = 0∧[(-1)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i20[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)

For Pair COND_LOAD731(TRUE, i19, i20) → LOAD731(+(i19, i20), +(i20, 2)) the following chains were created:
• We consider the chain LOAD731(i19[0], i20[0]) → COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0]), COND_LOAD731(TRUE, i19[1], i20[1]) → LOAD731(+(i19[1], i20[1]), +(i20[1], 2)), LOAD731(i19[0], i20[0]) → COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0]) which results in the following constraint:

We simplified constraint (7) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:

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

(9)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥)∧[(-1)bni_16 + (-1)Bound*bni_16] + [(-1)bni_16]i20[0] ≥ 0∧[2 + (-1)bso_17] ≥ 0)

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

(10)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥)∧[(-1)bni_16 + (-1)Bound*bni_16] + [(-1)bni_16]i20[0] ≥ 0∧[2 + (-1)bso_17] ≥ 0)

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

(11)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥)∧[(-1)bni_16 + (-1)Bound*bni_16] + [(-1)bni_16]i20[0] ≥ 0∧[2 + (-1)bso_17] ≥ 0)

We simplified constraint (11) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:

(12)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥)∧0 = 0∧[(-1)bni_16 + (-1)Bound*bni_16] + [(-1)bni_16]i20[0] ≥ 0∧0 = 0∧[2 + (-1)bso_17] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD731(i19, i20) → COND_LOAD731(&&(&&(>=(i20, 0), <=(i20, 1000)), >(+(i20, 2), 0)), i19, i20)
• (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥)∧0 = 0∧[(-1)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i20[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)

• (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥)∧0 = 0∧[(-1)bni_16 + (-1)Bound*bni_16] + [(-1)bni_16]i20[0] ≥ 0∧0 = 0∧[2 + (-1)bso_17] ≥ 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) = [2]
POL(LOAD731(x1, x2)) = [-1] + [-1]x2
POL(COND_LOAD731(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>:

The following pairs are in Pbound:

LOAD731(i19[0], i20[0]) → COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])

The following pairs are in P:

LOAD731(i19[0], i20[0]) → COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])

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

TRUE1&&(TRUE, TRUE)1
FALSE1&&(TRUE, FALSE)1
FALSE1&&(FALSE, TRUE)1
FALSE1&&(FALSE, FALSE)1

### (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:

Boolean, Integer

R is empty.

The integer pair graph contains the following rules and edges:
(0): LOAD731(i19[0], i20[0]) → COND_LOAD731(i20[0] >= 0 && i20[0] <= 1000 && i20[0] + 2 > 0, i19[0], i20[0])

The set Q consists of the following terms:

### (30) IDependencyGraphProof (EQUIVALENT transformation)

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

### (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:
none

R is empty.

The integer pair graph is empty.

The set Q consists of the following terms:

### (33) IDependencyGraphProof (EQUIVALENT transformation)

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

### (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:
Load404(1000, 1000, 1000, i4, i5) → Cond_Load404(i5 >= 0 && i5 <= 1000 && i5 + 1 > 0, 1000, 1000, 1000, i4, i5)
Cond_Load404(TRUE, 1000, 1000, 1000, i4, i5) → Load404(1000, 1000, 1000, i4 + i5, i5 + 1)
The set Q consists of the following terms:
Cond_Load404(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:
Load404(i4, i5) → Cond_Load404(i5 >= 0 && i5 <= 1000 && i5 + 1 > 0, i4, i5)
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:
Load404(i4, i5) → Cond_Load404(i5 >= 0 && i5 <= 1000 && i5 + 1 > 0, i4, i5)

The integer pair graph contains the following rules and edges:
(0): LOAD404(i4[0], i5[0]) → COND_LOAD404(i5[0] >= 0 && i5[0] <= 1000 && i5[0] + 1 > 0, i4[0], i5[0])

(0) -> (1), if ((i5[0] >= 0 && i5[0] <= 1000 && i5[0] + 1 > 0* TRUE)∧(i4[0]* i4[1])∧(i5[0]* i5[1]))

(1) -> (0), if ((i5[1] + 1* i5[0])∧(i4[1] + i5[1]* i4[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): LOAD404(i4[0], i5[0]) → COND_LOAD404(i5[0] >= 0 && i5[0] <= 1000 && i5[0] + 1 > 0, i4[0], i5[0])

(0) -> (1), if ((i5[0] >= 0 && i5[0] <= 1000 && i5[0] + 1 > 0* TRUE)∧(i4[0]* i4[1])∧(i5[0]* i5[1]))

(1) -> (0), if ((i5[1] + 1* i5[0])∧(i4[1] + i5[1]* i4[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 LOAD404(i4, i5) → COND_LOAD404(&&(&&(>=(i5, 0), <=(i5, 1000)), >(+(i5, 1), 0)), i4, i5) the following chains were created:
• We consider the chain LOAD404(i4[0], i5[0]) → COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0]), COND_LOAD404(TRUE, i4[1], i5[1]) → LOAD404(+(i4[1], i5[1]), +(i5[1], 1)) which results in the following constraint:

(1)    (&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0))=TRUEi4[0]=i4[1]i5[0]=i5[1]LOAD404(i4[0], i5[0])≥NonInfC∧LOAD404(i4[0], i5[0])≥COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])∧(UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥))

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

(2)    (>(+(i5[0], 1), 0)=TRUE>=(i5[0], 0)=TRUE<=(i5[0], 1000)=TRUELOAD404(i4[0], i5[0])≥NonInfC∧LOAD404(i4[0], i5[0])≥COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])∧(UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥))

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

(3)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i5[0] ≥ 0∧[(-1)bso_15] ≥ 0)

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

(4)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i5[0] ≥ 0∧[(-1)bso_15] ≥ 0)

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

(5)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i5[0] ≥ 0∧[(-1)bso_15] ≥ 0)

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

(6)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥)∧0 = 0∧[(-1)Bound*bni_14] + [(-1)bni_14]i5[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)

For Pair COND_LOAD404(TRUE, i4, i5) → LOAD404(+(i4, i5), +(i5, 1)) the following chains were created:
• We consider the chain LOAD404(i4[0], i5[0]) → COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0]), COND_LOAD404(TRUE, i4[1], i5[1]) → LOAD404(+(i4[1], i5[1]), +(i5[1], 1)), LOAD404(i4[0], i5[0]) → COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0]) which results in the following constraint:

We simplified constraint (7) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:

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

(9)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i5[0] ≥ 0∧[1 + (-1)bso_17] ≥ 0)

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

(10)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i5[0] ≥ 0∧[1 + (-1)bso_17] ≥ 0)

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

(11)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i5[0] ≥ 0∧[1 + (-1)bso_17] ≥ 0)

We simplified constraint (11) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:

(12)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥)∧0 = 0∧[(-1)Bound*bni_16] + [(-1)bni_16]i5[0] ≥ 0∧0 = 0∧[1 + (-1)bso_17] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD404(i4, i5) → COND_LOAD404(&&(&&(>=(i5, 0), <=(i5, 1000)), >(+(i5, 1), 0)), i4, i5)
• (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥)∧0 = 0∧[(-1)Bound*bni_14] + [(-1)bni_14]i5[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)

• (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥)∧0 = 0∧[(-1)Bound*bni_16] + [(-1)bni_16]i5[0] ≥ 0∧0 = 0∧[1 + (-1)bso_17] ≥ 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) = [2]
POL(FALSE) = [2]
POL(COND_LOAD404(x1, x2, x3)) = [2] + [-1]x3 + [-1]x1
POL(&&(x1, x2)) = [2]
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:

LOAD404(i4[0], i5[0]) → COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])

The following pairs are in P:

LOAD404(i4[0], i5[0]) → COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])

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

&&(TRUE, TRUE)1TRUE1
&&(TRUE, FALSE)1FALSE1
&&(FALSE, TRUE)1FALSE1
&&(FALSE, FALSE)1FALSE1

### (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): LOAD404(i4[0], i5[0]) → COND_LOAD404(i5[0] >= 0 && i5[0] <= 1000 && i5[0] + 1 > 0, i4[0], i5[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:
none

R is empty.

The integer pair graph is empty.

The set Q consists of the following terms: