(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: PastaB1

(1) JBC2FIG (SOUND transformation)

Constructed FIGraph.

(2) Obligation:

FIGraph based on JBC Program:
Graph of 162 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:
Load569(i57, i38) → Cond_Load569(i38 >= 0 && i57 > i38, i57, i38)
Cond_Load569(TRUE, i57, i38) → Load569(i57 + -1, i38)
The set Q consists of the following terms:
Load569(x0, x1)
Cond_Load569(TRUE, x0, x1)

(5) ITRStoIDPProof (EQUIVALENT transformation)

Added dependency pairs

(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:
Load569(i57, i38) → Cond_Load569(i38 >= 0 && i57 > i38, i57, i38)
Cond_Load569(TRUE, i57, i38) → Load569(i57 + -1, i38)

The integer pair graph contains the following rules and edges:
(0): LOAD569(i57[0], i38[0]) → COND_LOAD569(i38[0] >= 0 && i57[0] > i38[0], i57[0], i38[0])
(1): COND_LOAD569(TRUE, i57[1], i38[1]) → LOAD569(i57[1] + -1, i38[1])

(0) -> (1), if ((i57[0]* i57[1])∧(i38[0]* i38[1])∧(i38[0] >= 0 && i57[0] > i38[0]* TRUE))


(1) -> (0), if ((i38[1]* i38[0])∧(i57[1] + -1* i57[0]))



The set Q consists of the following terms:
Load569(x0, x1)
Cond_Load569(TRUE, x0, x1)

(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): LOAD569(i57[0], i38[0]) → COND_LOAD569(i38[0] >= 0 && i57[0] > i38[0], i57[0], i38[0])
(1): COND_LOAD569(TRUE, i57[1], i38[1]) → LOAD569(i57[1] + -1, i38[1])

(0) -> (1), if ((i57[0]* i57[1])∧(i38[0]* i38[1])∧(i38[0] >= 0 && i57[0] > i38[0]* TRUE))


(1) -> (0), if ((i38[1]* i38[0])∧(i57[1] + -1* i57[0]))



The set Q consists of the following terms:
Load569(x0, x1)
Cond_Load569(TRUE, x0, x1)

(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 LOAD569(i57, i38) → COND_LOAD569(&&(>=(i38, 0), >(i57, i38)), i57, i38) the following chains were created:
  • We consider the chain LOAD569(i57[0], i38[0]) → COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0]), COND_LOAD569(TRUE, i57[1], i38[1]) → LOAD569(+(i57[1], -1), i38[1]) which results in the following constraint:

    (1)    (i57[0]=i57[1]i38[0]=i38[1]&&(>=(i38[0], 0), >(i57[0], i38[0]))=TRUELOAD569(i57[0], i38[0])≥NonInfC∧LOAD569(i57[0], i38[0])≥COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])∧(UIncreasing(COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])), ≥))



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

    (2)    (>=(i38[0], 0)=TRUE>(i57[0], i38[0])=TRUELOAD569(i57[0], i38[0])≥NonInfC∧LOAD569(i57[0], i38[0])≥COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])∧(UIncreasing(COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])), ≥))



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

    (3)    (i38[0] ≥ 0∧i57[0] + [-1] + [-1]i38[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])), ≥)∧[(-1)Bound*bni_10] + [(-1)bni_10]i38[0] + [(2)bni_10]i57[0] ≥ 0∧[(-1)bso_11] ≥ 0)



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

    (4)    (i38[0] ≥ 0∧i57[0] + [-1] + [-1]i38[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])), ≥)∧[(-1)Bound*bni_10] + [(-1)bni_10]i38[0] + [(2)bni_10]i57[0] ≥ 0∧[(-1)bso_11] ≥ 0)



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

    (5)    (i38[0] ≥ 0∧i57[0] + [-1] + [-1]i38[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])), ≥)∧[(-1)Bound*bni_10] + [(-1)bni_10]i38[0] + [(2)bni_10]i57[0] ≥ 0∧[(-1)bso_11] ≥ 0)



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

    (6)    (i38[0] ≥ 0∧i57[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])), ≥)∧[(-1)Bound*bni_10 + (2)bni_10] + [bni_10]i38[0] + [(2)bni_10]i57[0] ≥ 0∧[(-1)bso_11] ≥ 0)







For Pair COND_LOAD569(TRUE, i57, i38) → LOAD569(+(i57, -1), i38) the following chains were created:
  • We consider the chain COND_LOAD569(TRUE, i57[1], i38[1]) → LOAD569(+(i57[1], -1), i38[1]) which results in the following constraint:

    (7)    (COND_LOAD569(TRUE, i57[1], i38[1])≥NonInfC∧COND_LOAD569(TRUE, i57[1], i38[1])≥LOAD569(+(i57[1], -1), i38[1])∧(UIncreasing(LOAD569(+(i57[1], -1), i38[1])), ≥))



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

    (8)    ((UIncreasing(LOAD569(+(i57[1], -1), i38[1])), ≥)∧[2 + (-1)bso_13] ≥ 0)



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

    (9)    ((UIncreasing(LOAD569(+(i57[1], -1), i38[1])), ≥)∧[2 + (-1)bso_13] ≥ 0)



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

    (10)    ((UIncreasing(LOAD569(+(i57[1], -1), i38[1])), ≥)∧[2 + (-1)bso_13] ≥ 0)



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

    (11)    ((UIncreasing(LOAD569(+(i57[1], -1), i38[1])), ≥)∧0 = 0∧0 = 0∧[2 + (-1)bso_13] ≥ 0)







To summarize, we get the following constraints P for the following pairs.
  • LOAD569(i57, i38) → COND_LOAD569(&&(>=(i38, 0), >(i57, i38)), i57, i38)
    • (i38[0] ≥ 0∧i57[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])), ≥)∧[(-1)Bound*bni_10 + (2)bni_10] + [bni_10]i38[0] + [(2)bni_10]i57[0] ≥ 0∧[(-1)bso_11] ≥ 0)

  • COND_LOAD569(TRUE, i57, i38) → LOAD569(+(i57, -1), i38)
    • ((UIncreasing(LOAD569(+(i57[1], -1), i38[1])), ≥)∧0 = 0∧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(LOAD569(x1, x2)) = [-1]x2 + [2]x1   
POL(COND_LOAD569(x1, x2, x3)) = [-1]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>:

COND_LOAD569(TRUE, i57[1], i38[1]) → LOAD569(+(i57[1], -1), i38[1])

The following pairs are in Pbound:

LOAD569(i57[0], i38[0]) → COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])

The following pairs are in P:

LOAD569(i57[0], i38[0]) → COND_LOAD569(&&(>=(i38[0], 0), >(i57[0], i38[0])), i57[0], i38[0])

There are no usable rules.

(10) Complex Obligation (AND)

(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): LOAD569(i57[0], i38[0]) → COND_LOAD569(i38[0] >= 0 && i57[0] > i38[0], i57[0], i38[0])


The set Q consists of the following terms:
Load569(x0, x1)
Cond_Load569(TRUE, x0, x1)

(12) IDependencyGraphProof (EQUIVALENT transformation)

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

(13) TRUE

(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_LOAD569(TRUE, i57[1], i38[1]) → LOAD569(i57[1] + -1, i38[1])


The set Q consists of the following terms:
Load569(x0, x1)
Cond_Load569(TRUE, x0, x1)

(15) IDependencyGraphProof (EQUIVALENT transformation)

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

(16) TRUE