### (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: PastaC7

### (1) JBC2FIG (SOUND transformation)

Constructed FIGraph.

### (2) Obligation:

FIGraph based on JBC Program:
Graph of 237 nodes with 1 SCC.

### (3) FIGtoITRSProof (SOUND transformation)

Transformed FIGraph to ITRS rules

### (4) 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:
Load1135(i101, i102, i103) → Cond_Load1135(i102 >= 0 && i102 <= i103 && i101 >= 0 && i101 <= 100 && i102 + 1 > 0, i101, i102, i103)
Cond_Load1135(TRUE, i101, i102, i103) → Load1135(i102, i102 + 1, i103 + -1)
The set Q consists of the following terms:

### (6) 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:
Load1135(i101, i102, i103) → Cond_Load1135(i102 >= 0 && i102 <= i103 && i101 >= 0 && i101 <= 100 && i102 + 1 > 0, i101, i102, i103)
Cond_Load1135(TRUE, i101, i102, i103) → Load1135(i102, i102 + 1, i103 + -1)

The integer pair graph contains the following rules and edges:
(0): LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(i102[0] >= 0 && i102[0] <= i103[0] && i101[0] >= 0 && i101[0] <= 100 && i102[0] + 1 > 0, i101[0], i102[0], i103[0])
(1): COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], i102[1] + 1, i103[1] + -1)

(0) -> (1), if ((i101[0]* i101[1])∧(i102[0]* i102[1])∧(i102[0] >= 0 && i102[0] <= i103[0] && i101[0] >= 0 && i101[0] <= 100 && i102[0] + 1 > 0* TRUE)∧(i103[0]* i103[1]))

(1) -> (0), if ((i102[1]* i101[0])∧(i103[1] + -1* i103[0])∧(i102[1] + 1* i102[0]))

The set Q consists of the following terms:

### (7) 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.

### (8) 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): LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(i102[0] >= 0 && i102[0] <= i103[0] && i101[0] >= 0 && i101[0] <= 100 && i102[0] + 1 > 0, i101[0], i102[0], i103[0])
(1): COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], i102[1] + 1, i103[1] + -1)

(0) -> (1), if ((i101[0]* i101[1])∧(i102[0]* i102[1])∧(i102[0] >= 0 && i102[0] <= i103[0] && i101[0] >= 0 && i101[0] <= 100 && i102[0] + 1 > 0* TRUE)∧(i103[0]* i103[1]))

(1) -> (0), if ((i102[1]* i101[0])∧(i103[1] + -1* i103[0])∧(i102[1] + 1* i102[0]))

The set Q consists of the following terms:

### (9) 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 LOAD1135(i101, i102, i103) → COND_LOAD1135(&&(&&(&&(&&(>=(i102, 0), <=(i102, i103)), >=(i101, 0)), <=(i101, 100)), >(+(i102, 1), 0)), i101, i102, i103) the following chains were created:
• We consider the chain LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0]), COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1)) which results in the following constraint:

(1)    (i101[0]=i101[1]i102[0]=i102[1]&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0))=TRUEi103[0]=i103[1]LOAD1135(i101[0], i102[0], i103[0])≥NonInfC∧LOAD1135(i101[0], i102[0], i103[0])≥COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])∧(UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥))

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

(2)    (>(+(i102[0], 1), 0)=TRUE<=(i101[0], 100)=TRUE>=(i101[0], 0)=TRUE>=(i102[0], 0)=TRUE<=(i102[0], i103[0])=TRUELOAD1135(i101[0], i102[0], i103[0])≥NonInfC∧LOAD1135(i101[0], i102[0], i103[0])≥COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])∧(UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥))

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

(3)    (i102[0] ≥ 0∧[100] + [-1]i101[0] ≥ 0∧i101[0] ≥ 0∧i102[0] ≥ 0∧i103[0] + [-1]i102[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥)∧[bni_12 + (-1)Bound*bni_12] + [bni_12]i103[0] + [(-1)bni_12]i102[0] ≥ 0∧[(-1)bso_13] ≥ 0)

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

(4)    (i102[0] ≥ 0∧[100] + [-1]i101[0] ≥ 0∧i101[0] ≥ 0∧i102[0] ≥ 0∧i103[0] + [-1]i102[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥)∧[bni_12 + (-1)Bound*bni_12] + [bni_12]i103[0] + [(-1)bni_12]i102[0] ≥ 0∧[(-1)bso_13] ≥ 0)

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

(5)    (i102[0] ≥ 0∧[100] + [-1]i101[0] ≥ 0∧i101[0] ≥ 0∧i102[0] ≥ 0∧i103[0] + [-1]i102[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥)∧[bni_12 + (-1)Bound*bni_12] + [bni_12]i103[0] + [(-1)bni_12]i102[0] ≥ 0∧[(-1)bso_13] ≥ 0)

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

(6)    (i102[0] ≥ 0∧[100] + [-1]i101[0] ≥ 0∧i101[0] ≥ 0∧i102[0] ≥ 0∧i103[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥)∧[bni_12 + (-1)Bound*bni_12] + [bni_12]i103[0] ≥ 0∧[(-1)bso_13] ≥ 0)

For Pair COND_LOAD1135(TRUE, i101, i102, i103) → LOAD1135(i102, +(i102, 1), +(i103, -1)) the following chains were created:
• We consider the chain COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], +(i102[1], 1), +(i103[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(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥)∧[2 + (-1)bso_15] ≥ 0)

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

(9)    ((UIncreasing(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥)∧[2 + (-1)bso_15] ≥ 0)

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

(10)    ((UIncreasing(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥)∧[2 + (-1)bso_15] ≥ 0)

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

(11)    ((UIncreasing(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥)∧0 = 0∧0 = 0∧0 = 0∧[2 + (-1)bso_15] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD1135(i101, i102, i103) → COND_LOAD1135(&&(&&(&&(&&(>=(i102, 0), <=(i102, i103)), >=(i101, 0)), <=(i101, 100)), >(+(i102, 1), 0)), i101, i102, i103)
• (i102[0] ≥ 0∧[100] + [-1]i101[0] ≥ 0∧i101[0] ≥ 0∧i102[0] ≥ 0∧i103[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥)∧[bni_12 + (-1)Bound*bni_12] + [bni_12]i103[0] ≥ 0∧[(-1)bso_13] ≥ 0)

• ((UIncreasing(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥)∧0 = 0∧0 = 0∧0 = 0∧[2 + (-1)bso_15] ≥ 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(LOAD1135(x1, x2, x3)) = [1] + x3 + [-1]x2
POL(COND_LOAD1135(x1, x2, x3, x4)) = [1] + x4 + [-1]x3
POL(&&(x1, x2)) = [-1]
POL(>=(x1, x2)) = [-1]
POL(0) = 0
POL(<=(x1, x2)) = [-1]
POL(100) = [100]
POL(>(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(1) = [1]
POL(-1) = [-1]

The following pairs are in P>:

The following pairs are in Pbound:

LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])

The following pairs are in P:

LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])

There are no usable rules.

### (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): LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(i102[0] >= 0 && i102[0] <= i103[0] && i101[0] >= 0 && i101[0] <= 100 && i102[0] + 1 > 0, i101[0], i102[0], i103[0])

The set Q consists of the following terms:

### (12) IDependencyGraphProof (EQUIVALENT transformation)

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

### (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:

Integer

R is empty.

The integer pair graph contains the following rules and edges:
(1): COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], i102[1] + 1, i103[1] + -1)

The set Q consists of the following terms: