(0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: PastaB7
/**
* Example taken from "A Term Rewriting Approach to the Automated Termination
* Analysis of Imperative Programs" (http://www.cs.unm.edu/~spf/papers/2009-02.pdf)
* and converted to Java.
*/

public class PastaB7 {
public static void main(String[] args) {
Random.args = args;
int x = Random.random();
int y = Random.random();
int z = Random.random();

while (x > z && y > z) {
x--;
y--;
}
}
}


public class Random {
static String[] args;
static int index = 0;

public static int random() {
String string = args[index];
index++;
return string.length();
}
}


(1) JBC2FIG (SOUND transformation)

Constructed FIGraph.

(2) Obligation:

FIGraph based on JBC Program:
Graph of 252 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:
Load1076(i8, i23, i59) → Cond_Load1076(i23 > i59 && i8 > i59, i8, i23, i59)
Cond_Load1076(TRUE, i8, i23, i59) → Load1076(i8 + -1, i23 + -1, i59)
The set Q consists of the following terms:
Load1076(x0, x1, x2)
Cond_Load1076(TRUE, x0, x1, x2)

(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:
Load1076(i8, i23, i59) → Cond_Load1076(i23 > i59 && i8 > i59, i8, i23, i59)
Cond_Load1076(TRUE, i8, i23, i59) → Load1076(i8 + -1, i23 + -1, i59)

The integer pair graph contains the following rules and edges:
(0): LOAD1076(i8[0], i23[0], i59[0]) → COND_LOAD1076(i23[0] > i59[0] && i8[0] > i59[0], i8[0], i23[0], i59[0])
(1): COND_LOAD1076(TRUE, i8[1], i23[1], i59[1]) → LOAD1076(i8[1] + -1, i23[1] + -1, i59[1])

(0) -> (1), if ((i8[0]* i8[1])∧(i23[0]* i23[1])∧(i23[0] > i59[0] && i8[0] > i59[0]* TRUE)∧(i59[0]* i59[1]))


(1) -> (0), if ((i8[1] + -1* i8[0])∧(i59[1]* i59[0])∧(i23[1] + -1* i23[0]))



The set Q consists of the following terms:
Load1076(x0, x1, x2)
Cond_Load1076(TRUE, x0, x1, x2)

(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): LOAD1076(i8[0], i23[0], i59[0]) → COND_LOAD1076(i23[0] > i59[0] && i8[0] > i59[0], i8[0], i23[0], i59[0])
(1): COND_LOAD1076(TRUE, i8[1], i23[1], i59[1]) → LOAD1076(i8[1] + -1, i23[1] + -1, i59[1])

(0) -> (1), if ((i8[0]* i8[1])∧(i23[0]* i23[1])∧(i23[0] > i59[0] && i8[0] > i59[0]* TRUE)∧(i59[0]* i59[1]))


(1) -> (0), if ((i8[1] + -1* i8[0])∧(i59[1]* i59[0])∧(i23[1] + -1* i23[0]))



The set Q consists of the following terms:
Load1076(x0, x1, x2)
Cond_Load1076(TRUE, x0, x1, x2)

(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 LOAD1076(i8, i23, i59) → COND_LOAD1076(&&(>(i23, i59), >(i8, i59)), i8, i23, i59) the following chains were created:
  • We consider the chain LOAD1076(i8[0], i23[0], i59[0]) → COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0]), COND_LOAD1076(TRUE, i8[1], i23[1], i59[1]) → LOAD1076(+(i8[1], -1), +(i23[1], -1), i59[1]) which results in the following constraint:

    (1)    (i8[0]=i8[1]i23[0]=i23[1]&&(>(i23[0], i59[0]), >(i8[0], i59[0]))=TRUEi59[0]=i59[1]LOAD1076(i8[0], i23[0], i59[0])≥NonInfC∧LOAD1076(i8[0], i23[0], i59[0])≥COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])∧(UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥))



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

    (2)    (>(i23[0], i59[0])=TRUE>(i8[0], i59[0])=TRUELOAD1076(i8[0], i23[0], i59[0])≥NonInfC∧LOAD1076(i8[0], i23[0], i59[0])≥COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])∧(UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥))



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

    (3)    (i23[0] + [-1] + [-1]i59[0] ≥ 0∧i8[0] + [-1] + [-1]i59[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥)∧[(-1)bni_10 + (-1)Bound*bni_10] + [(-1)bni_10]i59[0] + [bni_10]i23[0] ≥ 0∧[(-1)bso_11] ≥ 0)



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

    (4)    (i23[0] + [-1] + [-1]i59[0] ≥ 0∧i8[0] + [-1] + [-1]i59[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥)∧[(-1)bni_10 + (-1)Bound*bni_10] + [(-1)bni_10]i59[0] + [bni_10]i23[0] ≥ 0∧[(-1)bso_11] ≥ 0)



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

    (5)    (i23[0] + [-1] + [-1]i59[0] ≥ 0∧i8[0] + [-1] + [-1]i59[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥)∧[(-1)bni_10 + (-1)Bound*bni_10] + [(-1)bni_10]i59[0] + [bni_10]i23[0] ≥ 0∧[(-1)bso_11] ≥ 0)



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

    (6)    (i23[0] ≥ 0∧i8[0] + [-1] + [-1]i59[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥)∧[(-1)Bound*bni_10] + [bni_10]i23[0] ≥ 0∧[(-1)bso_11] ≥ 0)



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

    (7)    (i23[0] ≥ 0∧i59[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥)∧[(-1)Bound*bni_10] + [bni_10]i23[0] ≥ 0∧[(-1)bso_11] ≥ 0)



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

    (8)    (i23[0] ≥ 0∧i59[0] ≥ 0∧i8[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥)∧[(-1)Bound*bni_10] + [bni_10]i23[0] ≥ 0∧[(-1)bso_11] ≥ 0)


    (9)    (i23[0] ≥ 0∧i59[0] ≥ 0∧i8[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥)∧[(-1)Bound*bni_10] + [bni_10]i23[0] ≥ 0∧[(-1)bso_11] ≥ 0)







For Pair COND_LOAD1076(TRUE, i8, i23, i59) → LOAD1076(+(i8, -1), +(i23, -1), i59) the following chains were created:
  • We consider the chain COND_LOAD1076(TRUE, i8[1], i23[1], i59[1]) → LOAD1076(+(i8[1], -1), +(i23[1], -1), i59[1]) which results in the following constraint:

    (10)    (COND_LOAD1076(TRUE, i8[1], i23[1], i59[1])≥NonInfC∧COND_LOAD1076(TRUE, i8[1], i23[1], i59[1])≥LOAD1076(+(i8[1], -1), +(i23[1], -1), i59[1])∧(UIncreasing(LOAD1076(+(i8[1], -1), +(i23[1], -1), i59[1])), ≥))



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

    (11)    ((UIncreasing(LOAD1076(+(i8[1], -1), +(i23[1], -1), i59[1])), ≥)∧[1 + (-1)bso_13] ≥ 0)



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

    (12)    ((UIncreasing(LOAD1076(+(i8[1], -1), +(i23[1], -1), i59[1])), ≥)∧[1 + (-1)bso_13] ≥ 0)



    We simplified constraint (12) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:

    (13)    ((UIncreasing(LOAD1076(+(i8[1], -1), +(i23[1], -1), i59[1])), ≥)∧[1 + (-1)bso_13] ≥ 0)



    We simplified constraint (13) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:

    (14)    ((UIncreasing(LOAD1076(+(i8[1], -1), +(i23[1], -1), i59[1])), ≥)∧0 = 0∧0 = 0∧0 = 0∧[1 + (-1)bso_13] ≥ 0)







To summarize, we get the following constraints P for the following pairs.
  • LOAD1076(i8, i23, i59) → COND_LOAD1076(&&(>(i23, i59), >(i8, i59)), i8, i23, i59)
    • (i23[0] ≥ 0∧i59[0] ≥ 0∧i8[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥)∧[(-1)Bound*bni_10] + [bni_10]i23[0] ≥ 0∧[(-1)bso_11] ≥ 0)
    • (i23[0] ≥ 0∧i59[0] ≥ 0∧i8[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])), ≥)∧[(-1)Bound*bni_10] + [bni_10]i23[0] ≥ 0∧[(-1)bso_11] ≥ 0)

  • COND_LOAD1076(TRUE, i8, i23, i59) → LOAD1076(+(i8, -1), +(i23, -1), i59)
    • ((UIncreasing(LOAD1076(+(i8[1], -1), +(i23[1], -1), i59[1])), ≥)∧0 = 0∧0 = 0∧0 = 0∧[1 + (-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(LOAD1076(x1, x2, x3)) = [-1] + [-1]x3 + x2   
POL(COND_LOAD1076(x1, x2, x3, x4)) = [-1] + [-1]x4 + x3   
POL(&&(x1, x2)) = [-1]   
POL(>(x1, x2)) = [-1]   
POL(+(x1, x2)) = x1 + x2   
POL(-1) = [-1]   

The following pairs are in P>:

COND_LOAD1076(TRUE, i8[1], i23[1], i59[1]) → LOAD1076(+(i8[1], -1), +(i23[1], -1), i59[1])

The following pairs are in Pbound:

LOAD1076(i8[0], i23[0], i59[0]) → COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[0])

The following pairs are in P:

LOAD1076(i8[0], i23[0], i59[0]) → COND_LOAD1076(&&(>(i23[0], i59[0]), >(i8[0], i59[0])), i8[0], i23[0], i59[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): LOAD1076(i8[0], i23[0], i59[0]) → COND_LOAD1076(i23[0] > i59[0] && i8[0] > i59[0], i8[0], i23[0], i59[0])


The set Q consists of the following terms:
Load1076(x0, x1, x2)
Cond_Load1076(TRUE, x0, x1, x2)

(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_LOAD1076(TRUE, i8[1], i23[1], i59[1]) → LOAD1076(i8[1] + -1, i23[1] + -1, i59[1])


The set Q consists of the following terms:
Load1076(x0, x1, x2)
Cond_Load1076(TRUE, x0, x1, x2)

(15) IDependencyGraphProof (EQUIVALENT transformation)

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

(16) TRUE