(0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: PastaB2
`/** * 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 PastaB2 {    public static void main(String[] args) {        Random.args = args;        int x = Random.random();        int y = Random.random();        while (x > y) {            x--;            y++;        }    }}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:
PastaB2.main([Ljava/lang/String;)V: Graph of 179 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: PastaB2.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 8 rules for P and 0 rules for R.

P rules:
492_0_main_Load(EOS(STATIC_492), i90, i91, i90) → 495_0_main_LE(EOS(STATIC_495), i90, i91, i90, i91)
495_0_main_LE(EOS(STATIC_495), i90, i91, i90, i91) → 498_0_main_LE(EOS(STATIC_498), i90, i91, i90, i91)
498_0_main_LE(EOS(STATIC_498), i90, i91, i90, i91) → 502_0_main_Inc(EOS(STATIC_502), i90, i91) | >(i90, i91)
502_0_main_Inc(EOS(STATIC_502), i90, i91) → 507_0_main_Inc(EOS(STATIC_507), +(i90, -1), i91)
507_0_main_Inc(EOS(STATIC_507), i96, i91) → 509_0_main_JMP(EOS(STATIC_509), i96, +(i91, 1))
509_0_main_JMP(EOS(STATIC_509), i96, i97) → 524_0_main_Load(EOS(STATIC_524), i96, i97)
R rules:

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

P rules:
492_0_main_Load(EOS(STATIC_492), x0, x1, x0) → 492_0_main_Load(EOS(STATIC_492), +(x0, -1), +(x1, 1), +(x0, -1)) | <(x1, x0)
R rules:

Filtered ground terms:

EOS(x1) → EOS

Filtered duplicate args:

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

P rules:
R rules:

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

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

(1) -> (0), if (x1[1] + 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@67446579 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 492_0_MAIN_LOAD(x1, x0) → COND_492_0_MAIN_LOAD(<(x1, x0), x1, x0) the following chains were created:
• We consider the chain 492_0_MAIN_LOAD(x1[0], x0[0]) → COND_492_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0]), COND_492_0_MAIN_LOAD(TRUE, x1[1], x0[1]) → 492_0_MAIN_LOAD(+(x1[1], 1), +(x0[1], -1)) which results in the following constraint:

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

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_492_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)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_492_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)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_492_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)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_492_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] ≥ 0∧[(-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_492_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] ≥ 0∧[(-1)bso_9] ≥ 0)

(8)    (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(COND_492_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] ≥ 0∧[(-1)bso_9] ≥ 0)

For Pair COND_492_0_MAIN_LOAD(TRUE, x1, x0) → 492_0_MAIN_LOAD(+(x1, 1), +(x0, -1)) the following chains were created:
• We consider the chain COND_492_0_MAIN_LOAD(TRUE, x1[1], x0[1]) → 492_0_MAIN_LOAD(+(x1[1], 1), +(x0[1], -1)) which results in the following constraint:

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

(10)    ((UIncreasing(492_0_MAIN_LOAD(+(x1[1], 1), +(x0[1], -1))), ≥)∧[bni_10] = 0∧[2 + (-1)bso_11] ≥ 0)

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

(11)    ((UIncreasing(492_0_MAIN_LOAD(+(x1[1], 1), +(x0[1], -1))), ≥)∧[bni_10] = 0∧[2 + (-1)bso_11] ≥ 0)

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

(12)    ((UIncreasing(492_0_MAIN_LOAD(+(x1[1], 1), +(x0[1], -1))), ≥)∧[bni_10] = 0∧[2 + (-1)bso_11] ≥ 0)

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

(13)    ((UIncreasing(492_0_MAIN_LOAD(+(x1[1], 1), +(x0[1], -1))), ≥)∧[bni_10] = 0∧0 = 0∧0 = 0∧[2 + (-1)bso_11] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(COND_492_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] ≥ 0∧[(-1)bso_9] ≥ 0)
• (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(COND_492_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x0[0] ≥ 0∧[(-1)bso_9] ≥ 0)

• ((UIncreasing(492_0_MAIN_LOAD(+(x1[1], 1), +(x0[1], -1))), ≥)∧[bni_10] = 0∧0 = 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(492_0_MAIN_LOAD(x1, x2)) = [1] + x2 + [-1]x1
POL(COND_492_0_MAIN_LOAD(x1, x2, x3)) = [1] + x3 + [-1]x2
POL(<(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(1) = [1]
POL(-1) = [-1]

The following pairs are in P>:

The following pairs are in Pbound:

The following pairs are in P:

There are no usable rules.

(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:

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.

(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: