Termination of the given JBC could be proven:

(0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: example_2/Test

package example_2;


public class Test {

	public static int divBy(int x){
		int r = 0;
		int y;
		while (x > 0) {
			y = 2;
			x = x/y;
			r = r + x;
		}
		return r;
	}

	public static void main(String[] args) {
		if (args.length > 0) {
		        int x = args[0].length();
			int r = divBy(x);
			// System.out.println("Result: " + r);
		}
		// else System.out.println("Error: Incorrect call");
	}
}

(1) JBCToGraph (SOUND transformation)

Constructed TerminationGraph.

(2) Obligation:

Termination Graph based on JBC Program:
example_2.Test.main([Ljava/lang/String;)V: Graph of 88 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: example_2.Test.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 15 rules for P and 0 rules for R.

P rules:
197_0_divBy_LE(EOS(STATIC_197), i37, i37) → 202_0_divBy_LE(EOS(STATIC_202), i37, i37)
202_0_divBy_LE(EOS(STATIC_202), i37, i37) → 206_0_divBy_ConstantStackPush(EOS(STATIC_206), i37) | >(i37, 0)
206_0_divBy_ConstantStackPush(EOS(STATIC_206), i37) → 211_0_divBy_Store(EOS(STATIC_211), i37, 2)
211_0_divBy_Store(EOS(STATIC_211), i37, matching1) → 215_0_divBy_Load(EOS(STATIC_215), i37, 2) | =(matching1, 2)
215_0_divBy_Load(EOS(STATIC_215), i37, matching1) → 220_0_divBy_Load(EOS(STATIC_220), 2, i37) | =(matching1, 2)
220_0_divBy_Load(EOS(STATIC_220), matching1, i37) → 224_0_divBy_IntArithmetic(EOS(STATIC_224), i37, 2) | =(matching1, 2)
224_0_divBy_IntArithmetic(EOS(STATIC_224), i37, matching1) → 226_0_divBy_Store(EOS(STATIC_226), /(i37, 2)) | &&(>=(i37, 1), =(matching1, 2))
226_0_divBy_Store(EOS(STATIC_226), i39) → 228_0_divBy_Load(EOS(STATIC_228), i39)
228_0_divBy_Load(EOS(STATIC_228), i39) → 231_0_divBy_Load(EOS(STATIC_231), i39)
231_0_divBy_Load(EOS(STATIC_231), i39) → 233_0_divBy_IntArithmetic(EOS(STATIC_233), i39, i39)
233_0_divBy_IntArithmetic(EOS(STATIC_233), i39, i39) → 234_0_divBy_Store(EOS(STATIC_234), i39) | >=(i39, 0)
234_0_divBy_Store(EOS(STATIC_234), i39) → 237_0_divBy_JMP(EOS(STATIC_237), i39)
237_0_divBy_JMP(EOS(STATIC_237), i39) → 242_0_divBy_Load(EOS(STATIC_242), i39)
242_0_divBy_Load(EOS(STATIC_242), i39) → 192_0_divBy_Load(EOS(STATIC_192), i39)
192_0_divBy_Load(EOS(STATIC_192), i26) → 197_0_divBy_LE(EOS(STATIC_197), i26, i26)
R rules:

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

P rules:
197_0_divBy_LE(EOS(STATIC_197), x0, x0) → 197_0_divBy_LE(EOS(STATIC_197), /(x0, 2), /(x0, 2)) | &&(>(+(x0, 1), 1), <=(0, /(x0, 2)))
R rules:

Filtered ground terms:

197_0_divBy_LE(x1, x2, x3) → 197_0_divBy_LE(x2, x3)
EOS(x1) → EOS
Cond_197_0_divBy_LE(x1, x2, x3, x4) → Cond_197_0_divBy_LE(x1, x3, x4)

Filtered duplicate args:

197_0_divBy_LE(x1, x2) → 197_0_divBy_LE(x2)
Cond_197_0_divBy_LE(x1, x2, x3) → Cond_197_0_divBy_LE(x1, x3)

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

P rules:
197_0_divBy_LE(x0) → 197_0_divBy_LE(/(x0, 2)) | &&(>(x0, 0), <=(0, /(x0, 2)))
R rules:

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

P rules:
197_0_DIVBY_LE(x0) → COND_197_0_DIVBY_LE(&&(>(x0, 0), <=(0, /(x0, 2))), x0)
COND_197_0_DIVBY_LE(TRUE, x0) → 197_0_DIVBY_LE(/(x0, 2))
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:

Boolean, Integer

R is empty.

The integer pair graph contains the following rules and edges:
(0): 197_0_DIVBY_LE(x0[0]) → COND_197_0_DIVBY_LE(x0[0] > 0 && 0 <= x0[0] / 2, x0[0])
(1): COND_197_0_DIVBY_LE(TRUE, x0[1]) → 197_0_DIVBY_LE(x0[1] / 2)

(0) -> (1), if (x0[0] > 0 && 0 <= x0[0] / 2 ∧x0[0] →^* x0[1])

(1) -> (0), if (x0[1] / 2 →^* 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.IdpDefaultShapeHeuristic@ebe5687 Constraint Generator: NonInfConstraintGenerator: PathGenerator: MetricPathGenerator: Max Left Steps: 1 Max Right Steps: 1

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 197_0_DIVBY_LE(x0) → COND_197_0_DIVBY_LE(&&(>(x0, 0), <=(0, /(x0, 2))), x0) the following chains were created:

• We consider the chain 197_0_DIVBY_LE(x0[0]) → COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0]), COND_197_0_DIVBY_LE(TRUE, x0[1]) → 197_0_DIVBY_LE(/(x0[1], 2)) which results in the following constraint:
  

  (1)    (&&(>(x0[0], 0), <=(0, /(x0[0], 2)))=TRUE∧x0[0]=x0[1] ⇒ 197_0_DIVBY_LE(x0[0])≥NonInfC∧197_0_DIVBY_LE(x0[0])≥COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])∧(U^Increasing(COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])), ≥))

  
  
  We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint:
  

  (2)    (>(x0[0], 0)=TRUE∧<=(0, /(x0[0], 2))=TRUE ⇒ 197_0_DIVBY_LE(x0[0])≥NonInfC∧197_0_DIVBY_LE(x0[0])≥COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])∧(U^Increasing(COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])), ≥))

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

  (3)    (x0[0] + [-1] ≥ 0∧max{x0[0], [-1]x0[0]} + [-1] ≥ 0 ⇒ (U^Increasing(COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 0)

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

  (4)    (x0[0] + [-1] ≥ 0∧max{x0[0], [-1]x0[0]} + [-1] ≥ 0 ⇒ (U^Increasing(COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 0)

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

  (5)    (x0[0] + [-1] ≥ 0∧[2]x0[0] ≥ 0∧x0[0] + [-1] ≥ 0 ⇒ (U^Increasing(COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 0)

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

  (6)    (x0[0] ≥ 0∧[2] + [2]x0[0] ≥ 0∧x0[0] ≥ 0 ⇒ (U^Increasing(COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])), ≥)∧[(-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 0)

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

  (7)    (x0[0] ≥ 0∧x0[0] ≥ 0∧[1] + x0[0] ≥ 0 ⇒ (U^Increasing(COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])), ≥)∧[(-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 0)

  
  
  

For Pair COND_197_0_DIVBY_LE(TRUE, x0) → 197_0_DIVBY_LE(/(x0, 2)) the following chains were created:

• We consider the chain 197_0_DIVBY_LE(x0[0]) → COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0]), COND_197_0_DIVBY_LE(TRUE, x0[1]) → 197_0_DIVBY_LE(/(x0[1], 2)) which results in the following constraint:
  

  (8)    (&&(>(x0[0], 0), <=(0, /(x0[0], 2)))=TRUE∧x0[0]=x0[1] ⇒ COND_197_0_DIVBY_LE(TRUE, x0[1])≥NonInfC∧COND_197_0_DIVBY_LE(TRUE, x0[1])≥197_0_DIVBY_LE(/(x0[1], 2))∧(U^Increasing(197_0_DIVBY_LE(/(x0[1], 2))), ≥))

  
  
  We simplified constraint (8) using rules (III), (IDP_BOOLEAN) which results in the following new constraint:
  

  (9)    (>(x0[0], 0)=TRUE∧<=(0, /(x0[0], 2))=TRUE ⇒ COND_197_0_DIVBY_LE(TRUE, x0[0])≥NonInfC∧COND_197_0_DIVBY_LE(TRUE, x0[0])≥197_0_DIVBY_LE(/(x0[0], 2))∧(U^Increasing(197_0_DIVBY_LE(/(x0[1], 2))), ≥))

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

  (10)    (x0[0] + [-1] ≥ 0∧max{x0[0], [-1]x0[0]} + [-1] ≥ 0 ⇒ (U^Increasing(197_0_DIVBY_LE(/(x0[1], 2))), ≥)∧[(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[1 + (-1)bso_17] + x0[0] + [-1]max{x0[0], [-1]x0[0]} ≥ 0)

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

  (11)    (x0[0] + [-1] ≥ 0∧max{x0[0], [-1]x0[0]} + [-1] ≥ 0 ⇒ (U^Increasing(197_0_DIVBY_LE(/(x0[1], 2))), ≥)∧[(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[1 + (-1)bso_17] + x0[0] + [-1]max{x0[0], [-1]x0[0]} ≥ 0)

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

  (12)    (x0[0] + [-1] ≥ 0∧[2]x0[0] ≥ 0∧x0[0] + [-1] ≥ 0 ⇒ (U^Increasing(197_0_DIVBY_LE(/(x0[1], 2))), ≥)∧[(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[1 + (-1)bso_17] ≥ 0)

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

  (13)    (x0[0] ≥ 0∧[2] + [2]x0[0] ≥ 0∧x0[0] ≥ 0 ⇒ (U^Increasing(197_0_DIVBY_LE(/(x0[1], 2))), ≥)∧[(-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[1 + (-1)bso_17] ≥ 0)

  
  
  We simplified constraint (13) using rule (IDP_POLY_GCD) which results in the following new constraint:
  

  (14)    (x0[0] ≥ 0∧x0[0] ≥ 0∧[1] + x0[0] ≥ 0 ⇒ (U^Increasing(197_0_DIVBY_LE(/(x0[1], 2))), ≥)∧[(-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[1 + (-1)bso_17] ≥ 0)

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• 197_0_DIVBY_LE(x0) → COND_197_0_DIVBY_LE(&&(>(x0, 0), <=(0, /(x0, 2))), x0)
  
  • (x0[0] ≥ 0∧x0[0] ≥ 0∧[1] + x0[0] ≥ 0 ⇒ (U^Increasing(COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])), ≥)∧[(-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 0)
    
  
  
• COND_197_0_DIVBY_LE(TRUE, x0) → 197_0_DIVBY_LE(/(x0, 2))
  
  • (x0[0] ≥ 0∧x0[0] ≥ 0∧[1] + x0[0] ≥ 0 ⇒ (U^Increasing(197_0_DIVBY_LE(/(x0[1], 2))), ≥)∧[(-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[1 + (-1)bso_17] ≥ 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) = [1]   
POL(197_0_DIVBY_LE(x₁)) = [-1] + x₁   
POL(COND_197_0_DIVBY_LE(x₁, x₂)) = [-1] + x₂ + [-1]x₁   
POL(&&(x₁, x₂)) = 0   
POL(>(x₁, x₂)) = [-1]   
POL(0) = 0   
POL(<=(x₁, x₂)) = [-1]   
POL(2) = [2]   

Polynomial Interpretations with Context Sensitive Arithemetic Replacement
POL(Term^CSAR-Mode @ Context)

POL(/(x₁, 2)¹ @ {}) = max{x₁, [-1]x₁} + [-1]   
POL(/(x₁, 2)¹ @ {197_0_DIVBY_LE_1/0}) = max{x₁, [-1]x₁} + [-1]   

The following pairs are in P_>:

COND_197_0_DIVBY_LE(TRUE, x0[1]) → 197_0_DIVBY_LE(/(x0[1], 2))

The following pairs are in P_bound:

197_0_DIVBY_LE(x0[0]) → COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])
COND_197_0_DIVBY_LE(TRUE, x0[1]) → 197_0_DIVBY_LE(/(x0[1], 2))

The following pairs are in P_≥:

197_0_DIVBY_LE(x0[0]) → COND_197_0_DIVBY_LE(&&(>(x0[0], 0), <=(0, /(x0[0], 2))), x0[0])

At least the following rules have been oriented under context sensitive arithmetic replacement:

&&(TRUE, TRUE)¹ ↔ TRUE¹
FALSE¹ → &&(TRUE, FALSE)¹
FALSE¹ → &&(FALSE, TRUE)¹
FALSE¹ → &&(FALSE, FALSE)¹
/¹ →

(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:

Boolean, Integer

R is empty.

The integer pair graph contains the following rules and edges:
(0): 197_0_DIVBY_LE(x0[0]) → COND_197_0_DIVBY_LE(x0[0] > 0 && 0 <= x0[0] / 2, x0[0])


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