Termination of the given JBC could be proven:

0 JBC
↳1 JBC2FIG (⇐)
↳2 FIGraph
↳3 FIGtoITRSProof (⇐)
↳4 ITRS
↳5 ITRStoIDPProof (⇔)
↳6 IDP
↳7 UsableRulesProof (⇔)
↳8 IDP
↳9 IDPNonInfProof (⇐)
↳10 IDP
↳11 IDependencyGraphProof (⇔)
↳12 TRUE

(0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: PastaB1 
/**
 * 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 PastaB1 {
    public static void main(String[] args) {
        Random.args = args;
        int x = Random.random();
        int y = Random.random();

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


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 184 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:
Load599(i11, i26) → Cond_Load599(i11 > i26, i11, i26)
Cond_Load599(TRUE, i11, i26) → Load599(i11 + -1, i26)
The set Q consists of the following terms:
Load599(x0, x1)
Cond_Load599(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:

Integer


The ITRS R consists of the following rules:
Load599(i11, i26) → Cond_Load599(i11 > i26, i11, i26)
Cond_Load599(TRUE, i11, i26) → Load599(i11 + -1, i26)

The integer pair graph contains the following rules and edges:
(0): LOAD599(i11[0], i26[0]) → COND_LOAD599(i11[0] > i26[0], i11[0], i26[0])
(1): COND_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(i11[1] + -1, i26[1])

(0) -> (1), if ((i11[0] > i26[0] →* TRUE)∧(i11[0] →* i11[1])∧(i26[0] →* i26[1]))


(1) -> (0), if ((i11[1] + -1 →* i11[0])∧(i26[1] →* i26[0]))



The set Q consists of the following terms:
Load599(x0, x1)
Cond_Load599(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:

Integer


R is empty.

The integer pair graph contains the following rules and edges:
(0): LOAD599(i11[0], i26[0]) → COND_LOAD599(i11[0] > i26[0], i11[0], i26[0])
(1): COND_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(i11[1] + -1, i26[1])

(0) -> (1), if ((i11[0] > i26[0] →* TRUE)∧(i11[0] →* i11[1])∧(i26[0] →* i26[1]))


(1) -> (0), if ((i11[1] + -1 →* i11[0])∧(i26[1] →* i26[0]))



The set Q consists of the following terms:
Load599(x0, x1)
Cond_Load599(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 LOAD599(i11, i26) → COND_LOAD599(>(i11, i26), i11, i26) the following chains were created:
  • We consider the chain LOAD599(i11[0], i26[0]) → COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0]), COND_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(+(i11[1], -1), i26[1]) which results in the following constraint:

    (1)    (>(i11[0], i26[0])=TRUE∧i11[0]=i11[1]∧i26[0]=i26[1] ⇒ LOAD599(i11[0], i26[0])≥NonInfC∧LOAD599(i11[0], i26[0])≥COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])∧(UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥))



    We simplified constraint (1) using rule (IV) which results in the following new constraint:

    (2)    (>(i11[0], i26[0])=TRUE ⇒ LOAD599(i11[0], i26[0])≥NonInfC∧LOAD599(i11[0], i26[0])≥COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])∧(UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥))



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

    (3)    (i11[0] + [-1] + [-1]i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[bni_8 + (-1)Bound*bni_8] + [(-1)bni_8]i26[0] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)



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

    (4)    (i11[0] + [-1] + [-1]i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[bni_8 + (-1)Bound*bni_8] + [(-1)bni_8]i26[0] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)



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

    (5)    (i11[0] + [-1] + [-1]i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[bni_8 + (-1)Bound*bni_8] + [(-1)bni_8]i26[0] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)



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

    (6)    (i11[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)



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

    (7)    (i11[0] ≥ 0∧i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)


    (8)    (i11[0] ≥ 0∧i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)







For Pair COND_LOAD599(TRUE, i11, i26) → LOAD599(+(i11, -1), i26) the following chains were created:
  • We consider the chain COND_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(+(i11[1], -1), i26[1]) which results in the following constraint:

    (9)    (COND_LOAD599(TRUE, i11[1], i26[1])≥NonInfC∧COND_LOAD599(TRUE, i11[1], i26[1])≥LOAD599(+(i11[1], -1), i26[1])∧(UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥))



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

    (10)    ((UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥)∧[(-1)bso_11] ≥ 0)



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

    (11)    ((UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥)∧[(-1)bso_11] ≥ 0)



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

    (12)    ((UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥)∧[(-1)bso_11] ≥ 0)



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

    (13)    ((UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥)∧0 = 0∧0 = 0∧[(-1)bso_11] ≥ 0)







To summarize, we get the following constraints P≥ for the following pairs.
  • LOAD599(i11, i26) → COND_LOAD599(>(i11, i26), i11, i26)
    • (i11[0] ≥ 0∧i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)
    • (i11[0] ≥ 0∧i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)

  • COND_LOAD599(TRUE, i11, i26) → LOAD599(+(i11, -1), i26)
    • ((UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥)∧0 = 0∧0 = 0∧[(-1)bso_11] ≥ 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(LOAD599(x1, x2)) = [1] + [-1]x2 + x1   
POL(COND_LOAD599(x1, x2, x3)) = [-1]x3 + x2   
POL(>(x1, x2)) = [-1]   
POL(+(x1, x2)) = x1 + x2   
POL(-1) = [-1]   

The following pairs are in P>:

LOAD599(i11[0], i26[0]) → COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])

The following pairs are in Pbound:

LOAD599(i11[0], i26[0]) → COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])

The following pairs are in P≥:

COND_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(+(i11[1], -1), i26[1])

There are no usable rules.

(10) 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_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(i11[1] + -1, i26[1])


The set Q consists of the following terms:
Load599(x0, x1)
Cond_Load599(TRUE, x0, x1)

(11) IDependencyGraphProof (EQUIVALENT transformation)

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

(12) TRUE