### (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: PastaB7

### (1) JBC2FIG (SOUND transformation)

Constructed FIGraph.

### (2) Obligation:

FIGraph based on JBC Program:
Graph of 230 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:
Load921(i99, i100, i74) → Cond_Load921(i74 >= 0 && i100 > i74 && i99 > i74, i99, i100, i74)
Cond_Load921(TRUE, i99, i100, i74) → Load921(i99 + -1, i100 + -1, i74)
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:
Load921(i99, i100, i74) → Cond_Load921(i74 >= 0 && i100 > i74 && i99 > i74, i99, i100, i74)
Cond_Load921(TRUE, i99, i100, i74) → Load921(i99 + -1, i100 + -1, i74)

The integer pair graph contains the following rules and edges:
(0): LOAD921(i99[0], i100[0], i74[0]) → COND_LOAD921(i74[0] >= 0 && i100[0] > i74[0] && i99[0] > i74[0], i99[0], i100[0], i74[0])
(1): COND_LOAD921(TRUE, i99[1], i100[1], i74[1]) → LOAD921(i99[1] + -1, i100[1] + -1, i74[1])

(0) -> (1), if ((i74[0] >= 0 && i100[0] > i74[0] && i99[0] > i74[0]* TRUE)∧(i99[0]* i99[1])∧(i100[0]* i100[1])∧(i74[0]* i74[1]))

(1) -> (0), if ((i99[1] + -1* i99[0])∧(i74[1]* i74[0])∧(i100[1] + -1* i100[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): LOAD921(i99[0], i100[0], i74[0]) → COND_LOAD921(i74[0] >= 0 && i100[0] > i74[0] && i99[0] > i74[0], i99[0], i100[0], i74[0])
(1): COND_LOAD921(TRUE, i99[1], i100[1], i74[1]) → LOAD921(i99[1] + -1, i100[1] + -1, i74[1])

(0) -> (1), if ((i74[0] >= 0 && i100[0] > i74[0] && i99[0] > i74[0]* TRUE)∧(i99[0]* i99[1])∧(i100[0]* i100[1])∧(i74[0]* i74[1]))

(1) -> (0), if ((i99[1] + -1* i99[0])∧(i74[1]* i74[0])∧(i100[1] + -1* i100[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 LOAD921(i99, i100, i74) → COND_LOAD921(&&(&&(>=(i74, 0), >(i100, i74)), >(i99, i74)), i99, i100, i74) the following chains were created:
• We consider the chain LOAD921(i99[0], i100[0], i74[0]) → COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0]), COND_LOAD921(TRUE, i99[1], i100[1], i74[1]) → LOAD921(+(i99[1], -1), +(i100[1], -1), i74[1]) which results in the following constraint:

(1)    (&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0]))=TRUEi99[0]=i99[1]i100[0]=i100[1]i74[0]=i74[1]LOAD921(i99[0], i100[0], i74[0])≥NonInfC∧LOAD921(i99[0], i100[0], i74[0])≥COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])∧(UIncreasing(COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])), ≥))

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

(2)    (>(i99[0], i74[0])=TRUE>=(i74[0], 0)=TRUE>(i100[0], i74[0])=TRUELOAD921(i99[0], i100[0], i74[0])≥NonInfC∧LOAD921(i99[0], i100[0], i74[0])≥COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])∧(UIncreasing(COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])), ≥))

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

(3)    (i99[0] + [-1] + [-1]i74[0] ≥ 0∧i74[0] ≥ 0∧i100[0] + [-1] + [-1]i74[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])), ≥)∧[bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i74[0] + [(2)bni_11]i100[0] + [(2)bni_11]i99[0] ≥ 0∧[(-1)bso_12] ≥ 0)

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

(4)    (i99[0] + [-1] + [-1]i74[0] ≥ 0∧i74[0] ≥ 0∧i100[0] + [-1] + [-1]i74[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])), ≥)∧[bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i74[0] + [(2)bni_11]i100[0] + [(2)bni_11]i99[0] ≥ 0∧[(-1)bso_12] ≥ 0)

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

(5)    (i99[0] + [-1] + [-1]i74[0] ≥ 0∧i74[0] ≥ 0∧i100[0] + [-1] + [-1]i74[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])), ≥)∧[bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i74[0] + [(2)bni_11]i100[0] + [(2)bni_11]i99[0] ≥ 0∧[(-1)bso_12] ≥ 0)

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

(6)    (i99[0] ≥ 0∧i74[0] ≥ 0∧i100[0] + [-1] + [-1]i74[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])), ≥)∧[(3)bni_11 + (-1)Bound*bni_11] + [bni_11]i74[0] + [(2)bni_11]i100[0] + [(2)bni_11]i99[0] ≥ 0∧[(-1)bso_12] ≥ 0)

We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraint:

(7)    (i99[0] ≥ 0∧i74[0] ≥ 0∧i100[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])), ≥)∧[(5)bni_11 + (-1)Bound*bni_11] + [(3)bni_11]i74[0] + [(2)bni_11]i100[0] + [(2)bni_11]i99[0] ≥ 0∧[(-1)bso_12] ≥ 0)

For Pair COND_LOAD921(TRUE, i99, i100, i74) → LOAD921(+(i99, -1), +(i100, -1), i74) the following chains were created:
• We consider the chain COND_LOAD921(TRUE, i99[1], i100[1], i74[1]) → LOAD921(+(i99[1], -1), +(i100[1], -1), i74[1]) which results in the following constraint:

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

(9)    ((UIncreasing(LOAD921(+(i99[1], -1), +(i100[1], -1), i74[1])), ≥)∧[4 + (-1)bso_14] ≥ 0)

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

(10)    ((UIncreasing(LOAD921(+(i99[1], -1), +(i100[1], -1), i74[1])), ≥)∧[4 + (-1)bso_14] ≥ 0)

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

(11)    ((UIncreasing(LOAD921(+(i99[1], -1), +(i100[1], -1), i74[1])), ≥)∧[4 + (-1)bso_14] ≥ 0)

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

(12)    ((UIncreasing(LOAD921(+(i99[1], -1), +(i100[1], -1), i74[1])), ≥)∧0 = 0∧0 = 0∧0 = 0∧[4 + (-1)bso_14] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD921(i99, i100, i74) → COND_LOAD921(&&(&&(>=(i74, 0), >(i100, i74)), >(i99, i74)), i99, i100, i74)
• (i99[0] ≥ 0∧i74[0] ≥ 0∧i100[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])), ≥)∧[(5)bni_11 + (-1)Bound*bni_11] + [(3)bni_11]i74[0] + [(2)bni_11]i100[0] + [(2)bni_11]i99[0] ≥ 0∧[(-1)bso_12] ≥ 0)

• ((UIncreasing(LOAD921(+(i99[1], -1), +(i100[1], -1), i74[1])), ≥)∧0 = 0∧0 = 0∧0 = 0∧[4 + (-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(LOAD921(x1, x2, x3)) = [1] + [-1]x3 + [2]x2 + [2]x1
POL(COND_LOAD921(x1, x2, x3, x4)) = [1] + [-1]x4 + [2]x3 + [2]x2
POL(&&(x1, x2)) = [-1]
POL(>=(x1, x2)) = [-1]
POL(0) = 0
POL(>(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(-1) = [-1]

The following pairs are in P>:

The following pairs are in Pbound:

LOAD921(i99[0], i100[0], i74[0]) → COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[0])

The following pairs are in P:

LOAD921(i99[0], i100[0], i74[0]) → COND_LOAD921(&&(&&(>=(i74[0], 0), >(i100[0], i74[0])), >(i99[0], i74[0])), i99[0], i100[0], i74[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): LOAD921(i99[0], i100[0], i74[0]) → COND_LOAD921(i74[0] >= 0 && i100[0] > i74[0] && i99[0] > i74[0], i99[0], i100[0], i74[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_LOAD921(TRUE, i99[1], i100[1], i74[1]) → LOAD921(i99[1] + -1, i100[1] + -1, i74[1])

The set Q consists of the following terms: