### (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: PastaB5

### (1) JBC2FIG (SOUND transformation)

Constructed FIGraph.

### (2) Obligation:

FIGraph based on JBC Program:
Graph of 106 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:
Load139(i14) → Cond_Load139(i14 > 0 && 0 = i14 % 2, i14)
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:
Load139(i14) → Cond_Load139(i14 > 0 && 0 = i14 % 2, i14)

The integer pair graph contains the following rules and edges:
(0): LOAD139(i14[0]) → COND_LOAD139(i14[0] > 0 && 0 = i14[0] % 2, i14[0])
(1): COND_LOAD139(TRUE, i14[1]) → LOAD139(i14[1] + -1)

(0) -> (1), if ((i14[0] > 0 && 0 = i14[0] % 2* TRUE)∧(i14[0]* i14[1]))

(1) -> (0), if ((i14[1] + -1* i14[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): LOAD139(i14[0]) → COND_LOAD139(i14[0] > 0 && 0 = i14[0] % 2, i14[0])
(1): COND_LOAD139(TRUE, i14[1]) → LOAD139(i14[1] + -1)

(0) -> (1), if ((i14[0] > 0 && 0 = i14[0] % 2* TRUE)∧(i14[0]* i14[1]))

(1) -> (0), if ((i14[1] + -1* i14[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 LOAD139(i14) → COND_LOAD139(&&(>(i14, 0), =(0, %(i14, 2))), i14) the following chains were created:
• We consider the chain LOAD139(i14[0]) → COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0]), COND_LOAD139(TRUE, i14[1]) → LOAD139(+(i14[1], -1)) which results in the following constraint:

(1)    (&&(>(i14[0], 0), =(0, %(i14[0], 2)))=TRUEi14[0]=i14[1]LOAD139(i14[0])≥NonInfC∧LOAD139(i14[0])≥COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])∧(UIncreasing(COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])), ≥))

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

(2)    (>(i14[0], 0)=TRUE>=(0, %(i14[0], 2))=TRUE<=(0, %(i14[0], 2))=TRUELOAD139(i14[0])≥NonInfC∧LOAD139(i14[0])≥COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])∧(UIncreasing(COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])), ≥))

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

(3)    (i14[0] + [-1] ≥ 0∧[-1]min{[2], [-2]} ≥ 0∧max{[2], [-2]} ≥ 0 ⇒ (UIncreasing(COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])), ≥)∧[(-1)Bound*bni_10] + [(2)bni_10]i14[0] ≥ 0∧[(-1)bso_11] ≥ 0)

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

(4)    (i14[0] + [-1] ≥ 0∧[-1]min{[2], [-2]} ≥ 0∧max{[2], [-2]} ≥ 0 ⇒ (UIncreasing(COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])), ≥)∧[(-1)Bound*bni_10] + [(2)bni_10]i14[0] ≥ 0∧[(-1)bso_11] ≥ 0)

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

(5)    (i14[0] + [-1] ≥ 0∧[4] ≥ 0∧[2] ≥ 0∧[2] ≥ 0 ⇒ (UIncreasing(COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])), ≥)∧[(-1)Bound*bni_10] + [(2)bni_10]i14[0] ≥ 0∧[(-1)bso_11] ≥ 0)

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

(6)    (i14[0] ≥ 0∧[4] ≥ 0∧[2] ≥ 0∧[2] ≥ 0 ⇒ (UIncreasing(COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])), ≥)∧[(-1)Bound*bni_10 + (2)bni_10] + [(2)bni_10]i14[0] ≥ 0∧[(-1)bso_11] ≥ 0)

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

(7)    (i14[0] ≥ 0∧[1] ≥ 0∧[1] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])), ≥)∧[(-1)Bound*bni_10 + (2)bni_10] + [(2)bni_10]i14[0] ≥ 0∧[(-1)bso_11] ≥ 0)

For Pair COND_LOAD139(TRUE, i14) → LOAD139(+(i14, -1)) the following chains were created:
• We consider the chain COND_LOAD139(TRUE, i14[1]) → LOAD139(+(i14[1], -1)) which results in the following constraint:

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

(9)    ((UIncreasing(LOAD139(+(i14[1], -1))), ≥)∧[2 + (-1)bso_13] ≥ 0)

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

(10)    ((UIncreasing(LOAD139(+(i14[1], -1))), ≥)∧[2 + (-1)bso_13] ≥ 0)

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

(11)    ((UIncreasing(LOAD139(+(i14[1], -1))), ≥)∧[2 + (-1)bso_13] ≥ 0)

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

(12)    ((UIncreasing(LOAD139(+(i14[1], -1))), ≥)∧0 = 0∧[2 + (-1)bso_13] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD139(i14) → COND_LOAD139(&&(>(i14, 0), =(0, %(i14, 2))), i14)
• (i14[0] ≥ 0∧[1] ≥ 0∧[1] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])), ≥)∧[(-1)Bound*bni_10 + (2)bni_10] + [(2)bni_10]i14[0] ≥ 0∧[(-1)bso_11] ≥ 0)

• ((UIncreasing(LOAD139(+(i14[1], -1))), ≥)∧0 = 0∧[2 + (-1)bso_13] ≥ 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_LOAD139(x1, x2)) = [2]x2
POL(&&(x1, x2)) = [-1]
POL(>(x1, x2)) = [-1]
POL(0) = 0
POL(=(x1, x2)) = [-1]
POL(2) = [2]
POL(+(x1, x2)) = x1 + x2
POL(-1) = [-1]

Polynomial Interpretations with Context Sensitive Arithemetic Replacement
POL(TermCSAR-Mode @ Context)

POL(%(x1, 2)-1 @ {}) = min{x2, [-1]x2}
POL(%(x1, 2)1 @ {}) = max{x2, [-1]x2}

The following pairs are in P>:

The following pairs are in Pbound:

LOAD139(i14[0]) → COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[0])

The following pairs are in P:

LOAD139(i14[0]) → COND_LOAD139(&&(>(i14[0], 0), =(0, %(i14[0], 2))), i14[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): LOAD139(i14[0]) → COND_LOAD139(i14[0] > 0 && 0 = i14[0] % 2, i14[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_LOAD139(TRUE, i14[1]) → LOAD139(i14[1] + -1)

The set Q consists of the following terms: