### (0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: PastaB4
`/** * 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 PastaB4 {    public static void main(String[] args) {        Random.args = args;        int x = Random.random();        int y = Random.random();        while (x > y) {            int t = x;            x = y;            y = t;        }    }}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:
PastaB4.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: PastaB4.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 12 rules for P and 0 rules for R.

P rules:
292_0_main_Load(EOS(STATIC_292), i18, i46, i18) → 300_0_main_LE(EOS(STATIC_300), i18, i46, i18, i46)
300_0_main_LE(EOS(STATIC_300), i18, i46, i18, i46) → 314_0_main_LE(EOS(STATIC_314), i18, i46, i18, i46)
314_0_main_LE(EOS(STATIC_314), i18, i46, i18, i46) → 327_0_main_Load(EOS(STATIC_327), i18, i46) | >(i18, i46)
327_0_main_Load(EOS(STATIC_327), i18, i46) → 337_0_main_Store(EOS(STATIC_337), i46, i18)
337_0_main_Store(EOS(STATIC_337), i46, i18) → 345_0_main_Load(EOS(STATIC_345), i46, i18)
345_0_main_Load(EOS(STATIC_345), i46, i18) → 352_0_main_Store(EOS(STATIC_352), i18, i46)
352_0_main_Store(EOS(STATIC_352), i18, i46) → 359_0_main_Load(EOS(STATIC_359), i46, i18)
359_0_main_Load(EOS(STATIC_359), i46, i18) → 366_0_main_Store(EOS(STATIC_366), i46, i18)
366_0_main_Store(EOS(STATIC_366), i46, i18) → 375_0_main_JMP(EOS(STATIC_375), i46, i18)
375_0_main_JMP(EOS(STATIC_375), i46, i18) → 397_0_main_Load(EOS(STATIC_397), i46, i18)
R rules:

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

P rules:
292_0_main_Load(EOS(STATIC_292), x0, x1, x0) → 292_0_main_Load(EOS(STATIC_292), x1, x0, x1) | <(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 (x0[1]* x1[0]x1[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.IdpDefaultShapeHeuristic@1b710c15 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 292_0_MAIN_LOAD(x1, x0) → COND_292_0_MAIN_LOAD(<(x1, x0), x1, x0) the following chains were created:

We simplified constraint (1) using rules (III), (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_292_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x0[0] + [(-1)bni_11]x1[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] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(COND_292_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x0[0] + [(-1)bni_11]x1[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] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(COND_292_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x0[0] + [(-1)bni_11]x1[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 ⇒ (UIncreasing(COND_292_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 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_292_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 0)

(8)    (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(COND_292_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 0)

For Pair COND_292_0_MAIN_LOAD(TRUE, x1, x0) → 292_0_MAIN_LOAD(x0, x1) the following chains were created:

We simplified constraint (9) using rules (III), (IV) which results in the following new constraint:

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

(11)    (x0[0] + [-1] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(292_0_MAIN_LOAD(x0[1], x1[1])), ≥)∧[(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x0[0] + [(-1)bni_13]x1[0] ≥ 0∧[(-1)bso_14] + [2]x0[0] + [-2]x1[0] ≥ 0)

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

(12)    (x0[0] + [-1] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(292_0_MAIN_LOAD(x0[1], x1[1])), ≥)∧[(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x0[0] + [(-1)bni_13]x1[0] ≥ 0∧[(-1)bso_14] + [2]x0[0] + [-2]x1[0] ≥ 0)

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

(13)    (x0[0] + [-1] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(292_0_MAIN_LOAD(x0[1], x1[1])), ≥)∧[(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x0[0] + [(-1)bni_13]x1[0] ≥ 0∧[(-1)bso_14] + [2]x0[0] + [-2]x1[0] ≥ 0)

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

(14)    (x0[0] ≥ 0 ⇒ (UIncreasing(292_0_MAIN_LOAD(x0[1], x1[1])), ≥)∧[(-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[2 + (-1)bso_14] + [2]x0[0] ≥ 0)

We simplified constraint (14) using rule (IDP_SMT_SPLIT) which results in the following new constraints:

(15)    (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(292_0_MAIN_LOAD(x0[1], x1[1])), ≥)∧[(-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[2 + (-1)bso_14] + [2]x0[0] ≥ 0)

(16)    (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(292_0_MAIN_LOAD(x0[1], x1[1])), ≥)∧[(-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[2 + (-1)bso_14] + [2]x0[0] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(COND_292_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 0)
• (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(COND_292_0_MAIN_LOAD(<(x1[0], x0[0]), x1[0], x0[0])), ≥)∧[(-1)Bound*bni_11] + [bni_11]x0[0] ≥ 0∧[(-1)bso_12] ≥ 0)

• (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(292_0_MAIN_LOAD(x0[1], x1[1])), ≥)∧[(-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[2 + (-1)bso_14] + [2]x0[0] ≥ 0)
• (x0[0] ≥ 0∧x1[0] ≥ 0 ⇒ (UIncreasing(292_0_MAIN_LOAD(x0[1], x1[1])), ≥)∧[(-1)Bound*bni_13] + [bni_13]x0[0] ≥ 0∧[2 + (-1)bso_14] + [2]x0[0] ≥ 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) = [1]
POL(FALSE) = 0
POL(292_0_MAIN_LOAD(x1, x2)) = [-1] + x2 + [-1]x1
POL(COND_292_0_MAIN_LOAD(x1, x2, x3)) = [-1] + x3 + [-1]x2
POL(<(x1, x2)) = [-1]

The following pairs are in P>:

The following pairs are in Pbound:

The following pairs are in P:

There are no usable rules.

### (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:

Integer

R is empty.

The integer pair graph contains the following rules and edges: