(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) JBCToGraph (SOUND transformation)

Constructed TerminationGraph.

(2) Obligation:

Termination Graph based on JBC Program:
PastaB1.main([Ljava/lang/String;)V: Graph of 178 nodes with 1 SCC.


(3) TerminationGraphToSCCProof (SOUND transformation)

Splitted TerminationGraph to 1 SCCs.

(4) Obligation:

SCC of termination graph based on JBC Program.
SCC contains nodes from the following methods: PastaB1.main([Ljava/lang/String;)V
SCC calls the following helper methods:
Performed SCC analyses: UsedFieldsAnalysis

(5) SCCToIDPv1Proof (SOUND transformation)

Transformed FIGraph SCCs to IDPs. Log:

Generated 7 rules for P and 0 rules for R.


P rules:
532_0_main_Load(EOS(STATIC_532), i75, i46, i75) → 535_0_main_LE(EOS(STATIC_535), i75, i46, i75, i46)
535_0_main_LE(EOS(STATIC_535), i75, i46, i75, i46) → 538_0_main_LE(EOS(STATIC_538), i75, i46, i75, i46)
538_0_main_LE(EOS(STATIC_538), i75, i46, i75, i46) → 543_0_main_Inc(EOS(STATIC_543), i75, i46) | >(i75, i46)
543_0_main_Inc(EOS(STATIC_543), i75, i46) → 547_0_main_JMP(EOS(STATIC_547), +(i75, -1), i46)
547_0_main_JMP(EOS(STATIC_547), i78, i46) → 550_0_main_Load(EOS(STATIC_550), i78, i46)
550_0_main_Load(EOS(STATIC_550), i78, i46) → 530_0_main_Load(EOS(STATIC_530), i78, i46)
530_0_main_Load(EOS(STATIC_530), i75, i46) → 532_0_main_Load(EOS(STATIC_532), i75, i46, i75)
R rules:

Combined rules. Obtained 1 conditional rules for P and 0 conditional rules for R.


P rules:
532_0_main_Load(EOS(STATIC_532), x0, x1, x0) → 532_0_main_Load(EOS(STATIC_532), +(x0, -1), x1, +(x0, -1)) | <(x1, x0)
R rules:

Filtered ground terms:



532_0_main_Load(x1, x2, x3, x4) → 532_0_main_Load(x2, x3, x4)
EOS(x1) → EOS
Cond_532_0_main_Load(x1, x2, x3, x4, x5) → Cond_532_0_main_Load(x1, x3, x4, x5)

Filtered duplicate args:



532_0_main_Load(x1, x2, x3) → 532_0_main_Load(x2, x3)
Cond_532_0_main_Load(x1, x2, x3, x4) → Cond_532_0_main_Load(x1, x3, x4)

Combined rules. Obtained 1 conditional rules for P and 0 conditional rules for R.


P rules:
532_0_main_Load(x1, x0) → 532_0_main_Load(x1, +(x0, -1)) | <(x1, x0)
R rules:

Finished conversion. Obtained 2 rules for P and 0 rules for R. System has predefined symbols.


P rules:
532_0_MAIN_LOAD(x1, x0) → COND_532_0_MAIN_LOAD(<(x1, x0), x1, x0)
COND_532_0_MAIN_LOAD(TRUE, x1, x0) → 532_0_MAIN_LOAD(x1, +(x0, -1))
R rules:

(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


R is empty.

The integer pair graph contains the following rules and edges:
(0): 532_0_MAIN_LOAD(x1[0], x0[0]) → COND_532_0_MAIN_LOAD(x1[0] < x0[0], x1[0], x0[0])
(1): COND_532_0_MAIN_LOAD(TRUE, x1[1], x0[1]) → 532_0_MAIN_LOAD(x1[1], x0[1] + -1)

(0) -> (1), if (x1[0] < x0[0]x1[0]* x1[1]x0[0]* x0[1])


(1) -> (0), if (x1[1]* x1[0]x0[1] + -1* x0[0])



The set Q is empty.

(7) IDPNonInfProof (SOUND transformation)

Used the following options for this NonInfProof:
IDPGPoloSolver: Range: [(-1,2)] IsNat: false Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpCand1ShapeHeuristic@75e13ce3 Constraint Generator: NonInfConstraintGenerator: PathGenerator: MetricPathGenerator: Max Left Steps: 0 Max Right Steps: 0

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 532_0_MAIN_LOAD(x1, x0) → COND_532_0_MAIN_LOAD(<(x1, x0), x1, x0) the following chains were created:
  • We consider the chain 532_0_MAIN_LOAD(x1[0], x0[0]) → COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0]), COND_532_0_MAIN_LOAD(TRUE, x1[1], x0[1]) → 532_0_MAIN_LOAD(x1[1], +(x0[1], -1)) which results in the following constraint:

    (1)    (<(x1[0], x0[0])=TRUEx1[0]=x1[1]x0[0]=x0[1]532_0_MAIN_LOAD(x1[0], x0[0])≥NonInfC∧532_0_MAIN_LOAD(x1[0], x0[0])≥COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])∧(UIncreasing(COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥))



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

    (2)    (<(x1[0], x0[0])=TRUE532_0_MAIN_LOAD(x1[0], x0[0])≥NonInfC∧532_0_MAIN_LOAD(x1[0], x0[0])≥COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])∧(UIncreasing(COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥))



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

    (3)    (x0[0] + [-1] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] + [(-1)bni_8]x1[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)



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

    (4)    (x0[0] + [-1] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] + [(-1)bni_8]x1[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)    (x0[0] + [-1] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] + [(-1)bni_8]x1[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)



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

    (6)    (x0[0] ≥ 0 ⇒ (UIncreasing(COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)



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

    (7)    (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)


    (8)    (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)







For Pair COND_532_0_MAIN_LOAD(TRUE, x1, x0) → 532_0_MAIN_LOAD(x1, +(x0, -1)) the following chains were created:
  • We consider the chain COND_532_0_MAIN_LOAD(TRUE, x1[1], x0[1]) → 532_0_MAIN_LOAD(x1[1], +(x0[1], -1)) which results in the following constraint:

    (9)    (COND_532_0_MAIN_LOAD(TRUE, x1[1], x0[1])≥NonInfC∧COND_532_0_MAIN_LOAD(TRUE, x1[1], x0[1])≥532_0_MAIN_LOAD(x1[1], +(x0[1], -1))∧(UIncreasing(532_0_MAIN_LOAD(x1[1], +(x0[1], -1))), ≥))



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

    (10)    ((UIncreasing(532_0_MAIN_LOAD(x1[1], +(x0[1], -1))), ≥)∧[bni_10] = 0∧[(-1)bso_11] ≥ 0)



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

    (11)    ((UIncreasing(532_0_MAIN_LOAD(x1[1], +(x0[1], -1))), ≥)∧[bni_10] = 0∧[(-1)bso_11] ≥ 0)



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

    (12)    ((UIncreasing(532_0_MAIN_LOAD(x1[1], +(x0[1], -1))), ≥)∧[bni_10] = 0∧[(-1)bso_11] ≥ 0)



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

    (13)    ((UIncreasing(532_0_MAIN_LOAD(x1[1], +(x0[1], -1))), ≥)∧[bni_10] = 0∧0 = 0∧0 = 0∧[(-1)bso_11] ≥ 0)







To summarize, we get the following constraints P for the following pairs.
  • 532_0_MAIN_LOAD(x1, x0) → COND_532_0_MAIN_LOAD(<(x1, x0), x1, x0)
    • (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)
    • (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)

  • COND_532_0_MAIN_LOAD(TRUE, x1, x0) → 532_0_MAIN_LOAD(x1, +(x0, -1))
    • ((UIncreasing(532_0_MAIN_LOAD(x1[1], +(x0[1], -1))), ≥)∧[bni_10] = 0∧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(532_0_MAIN_LOAD(x1, x2)) = [1] + x2 + [-1]x1   
POL(COND_532_0_MAIN_LOAD(x1, x2, x3)) = x3 + [-1]x2   
POL(<(x1, x2)) = [-1]   
POL(+(x1, x2)) = x1 + x2   
POL(-1) = [-1]   

The following pairs are in P>:

532_0_MAIN_LOAD(x1[0], x0[0]) → COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])

The following pairs are in Pbound:

532_0_MAIN_LOAD(x1[0], x0[0]) → COND_532_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])

The following pairs are in P:

COND_532_0_MAIN_LOAD(TRUE, x1[1], x0[1]) → 532_0_MAIN_LOAD(x1[1], +(x0[1], -1))

There are no usable rules.

(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:
(1): COND_532_0_MAIN_LOAD(TRUE, x1[1], x0[1]) → 532_0_MAIN_LOAD(x1[1], x0[1] + -1)


The set Q is empty.

(9) IDependencyGraphProof (EQUIVALENT transformation)

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

(10) TRUE