(0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_25 (Sun Microsystems Inc.) Main-Class: ClassAnalysis/ClassAnalysis
package ClassAnalysis;

public class ClassAnalysis {
A field;

public static void main(String[] args) {
Random.args = args;
ClassAnalysis t = new ClassAnalysis();
t.field = new A();
t.field = new B();
t.eval();
}

public void eval() {
int x = Random.random() % 100;
this.field.test(x);
}
}

class A {
public boolean test(int x) {
while (x > 0) {
if (x > 10) {
x--;
} else {
x++;
}
}
return true;
}
}

class B extends A {
public boolean test(int x) {
while (x > 0) {
x--;
}
return true;
}
}


package ClassAnalysis;

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

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


(1) JBCToGraph (SOUND transformation)

Constructed TerminationGraph.

(2) Obligation:

Termination Graph based on JBC Program:
ClassAnalysis.ClassAnalysis.main([Ljava/lang/String;)V: Graph of 70 nodes with 0 SCCs.

ClassAnalysis.ClassAnalysis.eval()V: Graph of 86 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: ClassAnalysis.ClassAnalysis.eval()V
SCC calls the following helper methods:
Performed SCC analyses: UsedFieldsAnalysis

(5) SCCToIDPv1Proof (SOUND transformation)

Transformed FIGraph SCCs to IDPs. Log:

Generated 6 rules for P and 0 rules for R.


P rules:
343_0_test_LE(EOS(STATIC_343), i45, i45) → 359_0_test_LE(EOS(STATIC_359), i45, i45)
359_0_test_LE(EOS(STATIC_359), i45, i45) → 367_0_test_Inc(EOS(STATIC_367), i45) | >(i45, 0)
367_0_test_Inc(EOS(STATIC_367), i45) → 376_0_test_JMP(EOS(STATIC_376), +(i45, -1)) | >(i45, 0)
376_0_test_JMP(EOS(STATIC_376), i48) → 387_0_test_Load(EOS(STATIC_387), i48)
387_0_test_Load(EOS(STATIC_387), i48) → 331_0_test_Load(EOS(STATIC_331), i48)
331_0_test_Load(EOS(STATIC_331), i33) → 343_0_test_LE(EOS(STATIC_343), i33, i33)
R rules:

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


P rules:
343_0_test_LE(EOS(STATIC_343), x0, x0) → 343_0_test_LE(EOS(STATIC_343), +(x0, -1), +(x0, -1)) | >(x0, 0)
R rules:

Filtered ground terms:



343_0_test_LE(x1, x2, x3) → 343_0_test_LE(x2, x3)
EOS(x1) → EOS
Cond_343_0_test_LE(x1, x2, x3, x4) → Cond_343_0_test_LE(x1, x3, x4)

Filtered duplicate args:



343_0_test_LE(x1, x2) → 343_0_test_LE(x2)
Cond_343_0_test_LE(x1, x2, x3) → Cond_343_0_test_LE(x1, x3)

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


P rules:
343_0_test_LE(x0) → 343_0_test_LE(+(x0, -1)) | >(x0, 0)
R rules:

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


P rules:
343_0_TEST_LE(x0) → COND_343_0_TEST_LE(>(x0, 0), x0)
COND_343_0_TEST_LE(TRUE, x0) → 343_0_TEST_LE(+(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): 343_0_TEST_LE(x0[0]) → COND_343_0_TEST_LE(x0[0] > 0, x0[0])
(1): COND_343_0_TEST_LE(TRUE, x0[1]) → 343_0_TEST_LE(x0[1] + -1)

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


(1) -> (0), if (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@c324b85 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 343_0_TEST_LE(x0) → COND_343_0_TEST_LE(>(x0, 0), x0) the following chains were created:
  • We consider the chain 343_0_TEST_LE(x0[0]) → COND_343_0_TEST_LE(>(x0[0], 0), x0[0]), COND_343_0_TEST_LE(TRUE, x0[1]) → 343_0_TEST_LE(+(x0[1], -1)) which results in the following constraint:

    (1)    (>(x0[0], 0)=TRUEx0[0]=x0[1]343_0_TEST_LE(x0[0])≥NonInfC∧343_0_TEST_LE(x0[0])≥COND_343_0_TEST_LE(>(x0[0], 0), x0[0])∧(UIncreasing(COND_343_0_TEST_LE(>(x0[0], 0), x0[0])), ≥))



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

    (2)    (>(x0[0], 0)=TRUE343_0_TEST_LE(x0[0])≥NonInfC∧343_0_TEST_LE(x0[0])≥COND_343_0_TEST_LE(>(x0[0], 0), x0[0])∧(UIncreasing(COND_343_0_TEST_LE(>(x0[0], 0), x0[0])), ≥))



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

    (3)    (x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_343_0_TEST_LE(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_8] + [(2)bni_8]x0[0] ≥ 0∧[(-1)bso_9] ≥ 0)



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

    (4)    (x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_343_0_TEST_LE(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_8] + [(2)bni_8]x0[0] ≥ 0∧[(-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] ≥ 0 ⇒ (UIncreasing(COND_343_0_TEST_LE(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_8] + [(2)bni_8]x0[0] ≥ 0∧[(-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_343_0_TEST_LE(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_8 + (2)bni_8] + [(2)bni_8]x0[0] ≥ 0∧[(-1)bso_9] ≥ 0)







For Pair COND_343_0_TEST_LE(TRUE, x0) → 343_0_TEST_LE(+(x0, -1)) the following chains were created:
  • We consider the chain COND_343_0_TEST_LE(TRUE, x0[1]) → 343_0_TEST_LE(+(x0[1], -1)) which results in the following constraint:

    (7)    (COND_343_0_TEST_LE(TRUE, x0[1])≥NonInfC∧COND_343_0_TEST_LE(TRUE, x0[1])≥343_0_TEST_LE(+(x0[1], -1))∧(UIncreasing(343_0_TEST_LE(+(x0[1], -1))), ≥))



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

    (8)    ((UIncreasing(343_0_TEST_LE(+(x0[1], -1))), ≥)∧[bni_10] = 0∧[2 + (-1)bso_11] ≥ 0)



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

    (9)    ((UIncreasing(343_0_TEST_LE(+(x0[1], -1))), ≥)∧[bni_10] = 0∧[2 + (-1)bso_11] ≥ 0)



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

    (10)    ((UIncreasing(343_0_TEST_LE(+(x0[1], -1))), ≥)∧[bni_10] = 0∧[2 + (-1)bso_11] ≥ 0)



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

    (11)    ((UIncreasing(343_0_TEST_LE(+(x0[1], -1))), ≥)∧[bni_10] = 0∧0 = 0∧[2 + (-1)bso_11] ≥ 0)







To summarize, we get the following constraints P for the following pairs.
  • 343_0_TEST_LE(x0) → COND_343_0_TEST_LE(>(x0, 0), x0)
    • (x0[0] ≥ 0 ⇒ (UIncreasing(COND_343_0_TEST_LE(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_8 + (2)bni_8] + [(2)bni_8]x0[0] ≥ 0∧[(-1)bso_9] ≥ 0)

  • COND_343_0_TEST_LE(TRUE, x0) → 343_0_TEST_LE(+(x0, -1))
    • ((UIncreasing(343_0_TEST_LE(+(x0[1], -1))), ≥)∧[bni_10] = 0∧0 = 0∧[2 + (-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(343_0_TEST_LE(x1)) = [2]x1   
POL(COND_343_0_TEST_LE(x1, x2)) = [2]x2   
POL(>(x1, x2)) = [-1]   
POL(0) = 0   
POL(+(x1, x2)) = x1 + x2   
POL(-1) = [-1]   

The following pairs are in P>:

COND_343_0_TEST_LE(TRUE, x0[1]) → 343_0_TEST_LE(+(x0[1], -1))

The following pairs are in Pbound:

343_0_TEST_LE(x0[0]) → COND_343_0_TEST_LE(>(x0[0], 0), x0[0])

The following pairs are in P:

343_0_TEST_LE(x0[0]) → COND_343_0_TEST_LE(>(x0[0], 0), x0[0])

There are no usable rules.

(8) Complex Obligation (AND)

(9) 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): 343_0_TEST_LE(x0[0]) → COND_343_0_TEST_LE(x0[0] > 0, x0[0])


The set Q is empty.

(10) IDependencyGraphProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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_343_0_TEST_LE(TRUE, x0[1]) → 343_0_TEST_LE(x0[1] + -1)


The set Q is empty.

(13) IDependencyGraphProof (EQUIVALENT transformation)

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

(14) TRUE