### (0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: PastaA8
`/** * 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 PastaA8 {    public static void main(String[] args) {        Random.args = args;        int x = Random.random();        int y = Random.random();        while (x > y) {            x++;            y += 2;        }    }}public class Random {  static String[] args;  static int index = 0;  public static int random() {    String string = args[index];    index++;    return string.length();  }}`

### (1) JBC2FIG (SOUND transformation)

Constructed FIGraph.

### (2) Obligation:

FIGraph based on JBC Program:
PastaA8.main([Ljava/lang/String;)V: Graph of 154 nodes with 1 SCC.

### (3) FIGtoITRSProof (SOUND transformation)

Transformed FIGraph SCCs to IDPs. Logs:

Log for SCC 0:

Generated 8 rules for P and 2 rules for R.

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

Filtered ground terms:

Filtered duplicate args:

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

Finished conversion. Obtained 1 rules for P and 0 rules for R. System has predefined symbols.

### (4) 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): 551_0_MAIN_LOAD(x1[0], x0[0]) → COND_551_0_MAIN_LOAD(x1[0] >= 0 && x1[0] < x0[0] && x0[0] >= 0, x1[0], x0[0])

(0) -> (1), if ((x1[0] >= 0 && x1[0] < x0[0] && x0[0] >= 0* TRUE)∧(x1[0]* x1[1])∧(x0[0]* x0[1]))

(1) -> (0), if ((x1[1] + 2* x1[0])∧(x0[1] + 1* x0[0]))

The set Q is empty.

### (5) IDPNonInfProof (SOUND transformation)

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 551_0_MAIN_LOAD(x1, x0) → COND_551_0_MAIN_LOAD(&&(&&(>=(x1, 0), <(x1, x0)), >=(x0, 0)), x1, x0) the following chains were created:
• We consider the chain 551_0_MAIN_LOAD(x1[0], x0[0]) → COND_551_0_MAIN_LOAD(&&(&&(>=(x1[0], 0), <(x1[0], x0[0])), >=(x0[0], 0)), x1[0], x0[0]), COND_551_0_MAIN_LOAD(TRUE, x1[1], x0[1]) → 551_0_MAIN_LOAD(+(x1[1], 2), +(x0[1], 1)) which results in the following constraint:

(1)    (&&(&&(>=(x1[0], 0), <(x1[0], x0[0])), >=(x0[0], 0))=TRUEx1[0]=x1[1]x0[0]=x0[1]551_0_MAIN_LOAD(x1[0], x0[0])≥NonInfC∧551_0_MAIN_LOAD(x1[0], x0[0])≥COND_551_0_MAIN_LOAD(&&(&&(>=(x1[0], 0), <(x1[0], x0[0])), >=(x0[0], 0)), x1[0], x0[0])∧(UIncreasing(COND_551_0_MAIN_LOAD(&&(&&(>=(x1[0], 0), <(x1[0], x0[0])), >=(x0[0], 0)), x1[0], x0[0])), ≥))

We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) 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] ≥ 0∧x1[0] ≥ 0∧x0[0] + [-1] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(COND_551_0_MAIN_LOAD(&&(&&(>=(x1[0], 0), <(x1[0], x0[0])), >=(x0[0], 0)), x1[0], x0[0])), ≥)∧[(-1)bni_10 + (-1)Bound*bni_10] + [bni_10]x0[0] + [(-1)bni_10]x1[0] ≥ 0∧[(-1)bso_11] ≥ 0)

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

(4)    (x0[0] ≥ 0∧x1[0] ≥ 0∧x0[0] + [-1] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(COND_551_0_MAIN_LOAD(&&(&&(>=(x1[0], 0), <(x1[0], x0[0])), >=(x0[0], 0)), x1[0], x0[0])), ≥)∧[(-1)bni_10 + (-1)Bound*bni_10] + [bni_10]x0[0] + [(-1)bni_10]x1[0] ≥ 0∧[(-1)bso_11] ≥ 0)

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

(5)    (x0[0] ≥ 0∧x1[0] ≥ 0∧x0[0] + [-1] + [-1]x1[0] ≥ 0 ⇒ (UIncreasing(COND_551_0_MAIN_LOAD(&&(&&(>=(x1[0], 0), <(x1[0], x0[0])), >=(x0[0], 0)), x1[0], x0[0])), ≥)∧[(-1)bni_10 + (-1)Bound*bni_10] + [bni_10]x0[0] + [(-1)bni_10]x1[0] ≥ 0∧[(-1)bso_11] ≥ 0)

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

(6)    ([1] + x1[0] + x0[0] ≥ 0∧x1[0] ≥ 0∧x0[0] ≥ 0 ⇒ (UIncreasing(COND_551_0_MAIN_LOAD(&&(&&(>=(x1[0], 0), <(x1[0], x0[0])), >=(x0[0], 0)), x1[0], x0[0])), ≥)∧[(-1)Bound*bni_10] + [bni_10]x0[0] ≥ 0∧[(-1)bso_11] ≥ 0)

For Pair COND_551_0_MAIN_LOAD(TRUE, x1, x0) → 551_0_MAIN_LOAD(+(x1, 2), +(x0, 1)) the following chains were created:
• We consider the chain COND_551_0_MAIN_LOAD(TRUE, x1[1], x0[1]) → 551_0_MAIN_LOAD(+(x1[1], 2), +(x0[1], 1)) which results in the following constraint:

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

(8)    ((UIncreasing(551_0_MAIN_LOAD(+(x1[1], 2), +(x0[1], 1))), ≥)∧[1 + (-1)bso_13] ≥ 0)

We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:

(9)    ((UIncreasing(551_0_MAIN_LOAD(+(x1[1], 2), +(x0[1], 1))), ≥)∧[1 + (-1)bso_13] ≥ 0)

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

(10)    ((UIncreasing(551_0_MAIN_LOAD(+(x1[1], 2), +(x0[1], 1))), ≥)∧[1 + (-1)bso_13] ≥ 0)

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

(11)    ((UIncreasing(551_0_MAIN_LOAD(+(x1[1], 2), +(x0[1], 1))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_13] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• 551_0_MAIN_LOAD(x1, x0) → COND_551_0_MAIN_LOAD(&&(&&(>=(x1, 0), <(x1, x0)), >=(x0, 0)), x1, x0)
• ([1] + x1[0] + x0[0] ≥ 0∧x1[0] ≥ 0∧x0[0] ≥ 0 ⇒ (UIncreasing(COND_551_0_MAIN_LOAD(&&(&&(>=(x1[0], 0), <(x1[0], x0[0])), >=(x0[0], 0)), x1[0], x0[0])), ≥)∧[(-1)Bound*bni_10] + [bni_10]x0[0] ≥ 0∧[(-1)bso_11] ≥ 0)

• ((UIncreasing(551_0_MAIN_LOAD(+(x1[1], 2), +(x0[1], 1))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_13] ≥ 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(551_0_MAIN_LOAD(x1, x2)) = [-1] + x2 + [-1]x1
POL(COND_551_0_MAIN_LOAD(x1, x2, x3)) = [-1] + x3 + [-1]x2
POL(&&(x1, x2)) = [-1]
POL(>=(x1, x2)) = [-1]
POL(0) = 0
POL(<(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(2) = [2]
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.

### (7) 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): 551_0_MAIN_LOAD(x1[0], x0[0]) → COND_551_0_MAIN_LOAD(x1[0] >= 0 && x1[0] < x0[0] && x0[0] >= 0, x1[0], x0[0])

The set Q is empty.

### (8) IDependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

### (10) 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: