(0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_26 (Sun Microsystems Inc.) Main-Class: simple/whileDecr/Main
package simple.whileDecr;

public class Main {

/**
* @param args
*/
public static void main(String[] args) {
WhileDecr.decrease(args.length);
}

}


package simple.whileDecr;

public class WhileDecr {

// this example does acutally terminate.

public static void decrease(int i) {
while (i > 5) {
i--;
}
}
}


(1) JBCToGraph (SOUND transformation)

Constructed TerminationGraph.

(2) Obligation:

Termination Graph based on JBC Program:
simple.whileDecr.Main.main([Ljava/lang/String;)V: Graph of 39 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: simple.whileDecr.Main.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:
100_0_decrease_ConstantStackPush(EOS(STATIC_100), i12, i12) → 103_0_decrease_LE(EOS(STATIC_103), i12, i12, 5)
103_0_decrease_LE(EOS(STATIC_103), i24, i24, matching1) → 108_0_decrease_LE(EOS(STATIC_108), i24, i24, 5) | =(matching1, 5)
108_0_decrease_LE(EOS(STATIC_108), i24, i24, matching1) → 112_0_decrease_Inc(EOS(STATIC_112), i24) | &&(>(i24, 5), =(matching1, 5))
112_0_decrease_Inc(EOS(STATIC_112), i24) → 117_0_decrease_JMP(EOS(STATIC_117), +(i24, -1)) | >(i24, 0)
117_0_decrease_JMP(EOS(STATIC_117), i26) → 123_0_decrease_Load(EOS(STATIC_123), i26)
123_0_decrease_Load(EOS(STATIC_123), i26) → 96_0_decrease_Load(EOS(STATIC_96), i26)
96_0_decrease_Load(EOS(STATIC_96), i12) → 100_0_decrease_ConstantStackPush(EOS(STATIC_100), i12, i12)
R rules:

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


P rules:
100_0_decrease_ConstantStackPush(EOS(STATIC_100), x0, x0) → 100_0_decrease_ConstantStackPush(EOS(STATIC_100), +(x0, -1), +(x0, -1)) | >(x0, 5)
R rules:

Filtered ground terms:



100_0_decrease_ConstantStackPush(x1, x2, x3) → 100_0_decrease_ConstantStackPush(x2, x3)
EOS(x1) → EOS
Cond_100_0_decrease_ConstantStackPush(x1, x2, x3, x4) → Cond_100_0_decrease_ConstantStackPush(x1, x3, x4)

Filtered duplicate args:



100_0_decrease_ConstantStackPush(x1, x2) → 100_0_decrease_ConstantStackPush(x2)
Cond_100_0_decrease_ConstantStackPush(x1, x2, x3) → Cond_100_0_decrease_ConstantStackPush(x1, x3)

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


P rules:
100_0_decrease_ConstantStackPush(x0) → 100_0_decrease_ConstantStackPush(+(x0, -1)) | >(x0, 5)
R rules:

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


P rules:
100_0_DECREASE_CONSTANTSTACKPUSH(x0) → COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0, 5), x0)
COND_100_0_DECREASE_CONSTANTSTACKPUSH(TRUE, x0) → 100_0_DECREASE_CONSTANTSTACKPUSH(+(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): 100_0_DECREASE_CONSTANTSTACKPUSH(x0[0]) → COND_100_0_DECREASE_CONSTANTSTACKPUSH(x0[0] > 5, x0[0])
(1): COND_100_0_DECREASE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 100_0_DECREASE_CONSTANTSTACKPUSH(x0[1] + -1)

(0) -> (1), if (x0[0] > 5x0[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@2d68be1b 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 100_0_DECREASE_CONSTANTSTACKPUSH(x0) → COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0, 5), x0) the following chains were created:
  • We consider the chain 100_0_DECREASE_CONSTANTSTACKPUSH(x0[0]) → COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), x0[0]), COND_100_0_DECREASE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 100_0_DECREASE_CONSTANTSTACKPUSH(+(x0[1], -1)) which results in the following constraint:

    (1)    (>(x0[0], 5)=TRUEx0[0]=x0[1]100_0_DECREASE_CONSTANTSTACKPUSH(x0[0])≥NonInfC∧100_0_DECREASE_CONSTANTSTACKPUSH(x0[0])≥COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), x0[0])∧(UIncreasing(COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), x0[0])), ≥))



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

    (2)    (>(x0[0], 5)=TRUE100_0_DECREASE_CONSTANTSTACKPUSH(x0[0])≥NonInfC∧100_0_DECREASE_CONSTANTSTACKPUSH(x0[0])≥COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), x0[0])∧(UIncreasing(COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), x0[0])), ≥))



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

    (3)    (x0[0] + [-6] ≥ 0 ⇒ (UIncreasing(COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), 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] + [-6] ≥ 0 ⇒ (UIncreasing(COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), 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] + [-6] ≥ 0 ⇒ (UIncreasing(COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), 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_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), x0[0])), ≥)∧[(-1)Bound*bni_8 + (12)bni_8] + [(2)bni_8]x0[0] ≥ 0∧[(-1)bso_9] ≥ 0)







For Pair COND_100_0_DECREASE_CONSTANTSTACKPUSH(TRUE, x0) → 100_0_DECREASE_CONSTANTSTACKPUSH(+(x0, -1)) the following chains were created:
  • We consider the chain COND_100_0_DECREASE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 100_0_DECREASE_CONSTANTSTACKPUSH(+(x0[1], -1)) which results in the following constraint:

    (7)    (COND_100_0_DECREASE_CONSTANTSTACKPUSH(TRUE, x0[1])≥NonInfC∧COND_100_0_DECREASE_CONSTANTSTACKPUSH(TRUE, x0[1])≥100_0_DECREASE_CONSTANTSTACKPUSH(+(x0[1], -1))∧(UIncreasing(100_0_DECREASE_CONSTANTSTACKPUSH(+(x0[1], -1))), ≥))



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

    (8)    ((UIncreasing(100_0_DECREASE_CONSTANTSTACKPUSH(+(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(100_0_DECREASE_CONSTANTSTACKPUSH(+(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(100_0_DECREASE_CONSTANTSTACKPUSH(+(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(100_0_DECREASE_CONSTANTSTACKPUSH(+(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.
  • 100_0_DECREASE_CONSTANTSTACKPUSH(x0) → COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0, 5), x0)
    • (x0[0] ≥ 0 ⇒ (UIncreasing(COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), x0[0])), ≥)∧[(-1)Bound*bni_8 + (12)bni_8] + [(2)bni_8]x0[0] ≥ 0∧[(-1)bso_9] ≥ 0)

  • COND_100_0_DECREASE_CONSTANTSTACKPUSH(TRUE, x0) → 100_0_DECREASE_CONSTANTSTACKPUSH(+(x0, -1))
    • ((UIncreasing(100_0_DECREASE_CONSTANTSTACKPUSH(+(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(100_0_DECREASE_CONSTANTSTACKPUSH(x1)) = [2]x1   
POL(COND_100_0_DECREASE_CONSTANTSTACKPUSH(x1, x2)) = [2]x2   
POL(>(x1, x2)) = [-1]   
POL(5) = [5]   
POL(+(x1, x2)) = x1 + x2   
POL(-1) = [-1]   

The following pairs are in P>:

COND_100_0_DECREASE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 100_0_DECREASE_CONSTANTSTACKPUSH(+(x0[1], -1))

The following pairs are in Pbound:

100_0_DECREASE_CONSTANTSTACKPUSH(x0[0]) → COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), x0[0])

The following pairs are in P:

100_0_DECREASE_CONSTANTSTACKPUSH(x0[0]) → COND_100_0_DECREASE_CONSTANTSTACKPUSH(>(x0[0], 5), 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): 100_0_DECREASE_CONSTANTSTACKPUSH(x0[0]) → COND_100_0_DECREASE_CONSTANTSTACKPUSH(x0[0] > 5, 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_100_0_DECREASE_CONSTANTSTACKPUSH(TRUE, x0[1]) → 100_0_DECREASE_CONSTANTSTACKPUSH(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