Termination proof of /tmp/tmpYkbbKL/ClassAnalysis.jar

Termination of the given JBC could be proven:

0 JBC

↳1 JBCToGraph (⇒, 100 ms)

↳2 JBCTerminationGraph

↳3 TerminationGraphToSCCProof (⇒, 0 ms)

↳4 JBCTerminationSCC

↳5 SCCToIDPv1Proof (⇒, 20 ms)

↳6 IDP

↳7 IDPNonInfProof (⇒, 120 ms)

↳8 AND

    ↳9 IDP

        ↳10 IDependencyGraphProof (⇔, 0 ms)

        ↳11 TRUE

    ↳12 IDP

        ↳13 IDependencyGraphProof (⇔, 0 ms)

        ↳14 TRUE

(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:
335_0_test_LE(EOS(STATIC_335), i44, i44) → 352_0_test_LE(EOS(STATIC_352), i44, i44)
352_0_test_LE(EOS(STATIC_352), i44, i44) → 360_0_test_Inc(EOS(STATIC_360), i44) | >(i44, 0)
360_0_test_Inc(EOS(STATIC_360), i44) → 369_0_test_JMP(EOS(STATIC_369), +(i44, -1)) | >(i44, 0)
369_0_test_JMP(EOS(STATIC_369), i47) → 381_0_test_Load(EOS(STATIC_381), i47)
381_0_test_Load(EOS(STATIC_381), i47) → 322_0_test_Load(EOS(STATIC_322), i47)
322_0_test_Load(EOS(STATIC_322), i30) → 335_0_test_LE(EOS(STATIC_335), i30, i30)
R rules:

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

P rules:
335_0_test_LE(EOS(STATIC_335), x0, x0) → 335_0_test_LE(EOS(STATIC_335), +(x0, -1), +(x0, -1)) | >(x0, 0)
R rules:

Filtered ground terms:

335_0_test_LE(x1, x2, x3) → 335_0_test_LE(x2, x3)
EOS(x1) → EOS
Cond_335_0_test_LE(x1, x2, x3, x4) → Cond_335_0_test_LE(x1, x3, x4)

Filtered duplicate args:

335_0_test_LE(x1, x2) → 335_0_test_LE(x2)
Cond_335_0_test_LE(x1, x2, x3) → Cond_335_0_test_LE(x1, x3)

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

P rules:
335_0_test_LE(x0) → 335_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:
335_0_TEST_LE(x0) → COND_335_0_TEST_LE(>(x0, 0), x0)
COND_335_0_TEST_LE(TRUE, x0) → 335_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): 335_0_TEST_LE(x0[0]) → COND_335_0_TEST_LE(x0[0] > 0, x0[0])
(1): COND_335_0_TEST_LE(TRUE, x0[1]) → 335_0_TEST_LE(x0[1] + -1)

(0) -> (1), if (x0[0] > 0 ∧x0[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@6d0cecb2 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 335_0_TEST_LE(x0) → COND_335_0_TEST_LE(>(x0, 0), x0) the following chains were created:

We consider the chain 335_0_TEST_LE(x0[0]) → COND_335_0_TEST_LE(>(x0[0], 0), x0[0]), COND_335_0_TEST_LE(TRUE, x0[1]) → 335_0_TEST_LE(+(x0[1], -1)) which results in the following constraint:
(1)    (>(x0[0], 0)=TRUE∧x0[0]=x0[1] ⇒ 335_0_TEST_LE(x0[0])≥NonInfC∧335_0_TEST_LE(x0[0])≥COND_335_0_TEST_LE(>(x0[0], 0), x0[0])∧(U^Increasing(COND_335_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)=TRUE ⇒ 335_0_TEST_LE(x0[0])≥NonInfC∧335_0_TEST_LE(x0[0])≥COND_335_0_TEST_LE(>(x0[0], 0), x0[0])∧(U^Increasing(COND_335_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 ⇒ (U^Increasing(COND_335_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 ⇒ (U^Increasing(COND_335_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 ⇒ (U^Increasing(COND_335_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 ⇒ (U^Increasing(COND_335_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_335_0_TEST_LE(TRUE, x0) → 335_0_TEST_LE(+(x0, -1)) the following chains were created:

We consider the chain COND_335_0_TEST_LE(TRUE, x0[1]) → 335_0_TEST_LE(+(x0[1], -1)) which results in the following constraint:
(7)    (COND_335_0_TEST_LE(TRUE, x0[1])≥NonInfC∧COND_335_0_TEST_LE(TRUE, x0[1])≥335_0_TEST_LE(+(x0[1], -1))∧(U^Increasing(335_0_TEST_LE(+(x0[1], -1))), ≥))

We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(8)    ((U^Increasing(335_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)    ((U^Increasing(335_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)    ((U^Increasing(335_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)    ((U^Increasing(335_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.

335_0_TEST_LE(x0) → COND_335_0_TEST_LE(>(x0, 0), x0)
- (x0[0] ≥ 0 ⇒ (U^Increasing(COND_335_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_335_0_TEST_LE(TRUE, x0) → 335_0_TEST_LE(+(x0, -1))
- ((U^Increasing(335_0_TEST_LE(+(x0[1], -1))), ≥)∧[bni_10] = 0∧0 = 0∧[2 + (-1)bso_11] ≥ 0)

The constraints for P_> respective P_bound 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 P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:

POL(TRUE) = 0
POL(FALSE) = 0
POL(335_0_TEST_LE(x₁)) = [2]x₁
POL(COND_335_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_335_0_TEST_LE(TRUE, x0[1]) → 335_0_TEST_LE(+(x0[1], -1))

The following pairs are in P_bound:

335_0_TEST_LE(x0[0]) → COND_335_0_TEST_LE(>(x0[0], 0), x0[0])

The following pairs are in P_≥:

335_0_TEST_LE(x0[0]) → COND_335_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): 335_0_TEST_LE(x0[0]) → COND_335_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_335_0_TEST_LE(TRUE, x0[1]) → 335_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.

(0) Obligation:

(1) JBCToGraph (SOUND transformation)

(2) Obligation:

(3) TerminationGraphToSCCProof (SOUND transformation)

(4) Obligation:

(5) SCCToIDPv1Proof (SOUND transformation)

(6) Obligation:

(7) IDPNonInfProof (SOUND transformation)

(8) Complex Obligation (AND)

(9) Obligation:

(10) IDependencyGraphProof (EQUIVALENT transformation)

(11) TRUE

(12) Obligation:

(13) IDependencyGraphProof (EQUIVALENT transformation)

(14) TRUE