### (0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: PastaB7
`/** * 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 PastaB7 {    public static void main(String[] args) {        Random.args = args;        int x = Random.random();        int y = Random.random();        int z = Random.random();        while (x > z && y > z) {            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) JBC2FIG (SOUND transformation)

Constructed FIGraph.

### (2) Obligation:

FIGraph based on JBC Program:
PastaB7.main([Ljava/lang/String;)V: Graph of 230 nodes with 1 SCC.

### (3) FIGtoITRSProof (SOUND transformation)

Transformed FIGraph SCCs to IDPs. Logs:

Log for SCC 0:

Generated 12 rules for P and 5 rules for R.

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

Filtered ground terms:

Cond_1253_0_main_Load(x1, x2, x3, x4, x5, x6) → Cond_1253_0_main_Load(x1, x3, x4, x5, x6)

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): 1253_0_MAIN_LOAD(x1[0], x2[0], x0[0]) → COND_1253_0_MAIN_LOAD(x2[0] < x1[0] && x2[0] < x0[0], x1[0], x2[0], x0[0])
(1): COND_1253_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1]) → 1253_0_MAIN_LOAD(x1[1] + -1, x2[1], x0[1] + -1)

(0) -> (1), if ((x2[0] < x1[0] && x2[0] < x0[0]* TRUE)∧(x1[0]* x1[1])∧(x2[0]* x2[1])∧(x0[0]* x0[1]))

(1) -> (0), if ((x1[1] + -1* x1[0])∧(x2[1]* x2[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 1253_0_MAIN_LOAD(x1, x2, x0) → COND_1253_0_MAIN_LOAD(&&(<(x2, x1), <(x2, x0)), x1, x2, x0) the following chains were created:
• We consider the chain 1253_0_MAIN_LOAD(x1[0], x2[0], x0[0]) → COND_1253_0_MAIN_LOAD(&&(<(x2[0], x1[0]), <(x2[0], x0[0])), x1[0], x2[0], x0[0]), COND_1253_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1]) → 1253_0_MAIN_LOAD(+(x1[1], -1), x2[1], +(x0[1], -1)) which results in the following constraint:

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)    (x1[0] + [-1] + [-1]x2[0] ≥ 0∧x0[0] + [-1] + [-1]x2[0] ≥ 0 ⇒ (UIncreasing(COND_1253_0_MAIN_LOAD(&&(<(x2[0], x1[0]), <(x2[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(-1)Bound*bni_10] + [(-1)bni_10]x2[0] + [bni_10]x1[0] ≥ 0∧[1 + (-1)bso_11] ≥ 0)

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

(4)    (x1[0] + [-1] + [-1]x2[0] ≥ 0∧x0[0] + [-1] + [-1]x2[0] ≥ 0 ⇒ (UIncreasing(COND_1253_0_MAIN_LOAD(&&(<(x2[0], x1[0]), <(x2[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(-1)Bound*bni_10] + [(-1)bni_10]x2[0] + [bni_10]x1[0] ≥ 0∧[1 + (-1)bso_11] ≥ 0)

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

(5)    (x1[0] + [-1] + [-1]x2[0] ≥ 0∧x0[0] + [-1] + [-1]x2[0] ≥ 0 ⇒ (UIncreasing(COND_1253_0_MAIN_LOAD(&&(<(x2[0], x1[0]), <(x2[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(-1)Bound*bni_10] + [(-1)bni_10]x2[0] + [bni_10]x1[0] ≥ 0∧[1 + (-1)bso_11] ≥ 0)

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

(6)    (x1[0] ≥ 0∧x0[0] + [-1] + [-1]x2[0] ≥ 0 ⇒ (UIncreasing(COND_1253_0_MAIN_LOAD(&&(<(x2[0], x1[0]), <(x2[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(-1)Bound*bni_10 + bni_10] + [bni_10]x1[0] ≥ 0∧[1 + (-1)bso_11] ≥ 0)

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

(7)    (x1[0] ≥ 0∧x2[0] ≥ 0 ⇒ (UIncreasing(COND_1253_0_MAIN_LOAD(&&(<(x2[0], x1[0]), <(x2[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(-1)Bound*bni_10 + bni_10] + [bni_10]x1[0] ≥ 0∧[1 + (-1)bso_11] ≥ 0)

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

(8)    (x1[0] ≥ 0∧x2[0] ≥ 0∧x0[0] ≥ 0 ⇒ (UIncreasing(COND_1253_0_MAIN_LOAD(&&(<(x2[0], x1[0]), <(x2[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(-1)Bound*bni_10 + bni_10] + [bni_10]x1[0] ≥ 0∧[1 + (-1)bso_11] ≥ 0)

(9)    (x1[0] ≥ 0∧x2[0] ≥ 0∧x0[0] ≥ 0 ⇒ (UIncreasing(COND_1253_0_MAIN_LOAD(&&(<(x2[0], x1[0]), <(x2[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(-1)Bound*bni_10 + bni_10] + [bni_10]x1[0] ≥ 0∧[1 + (-1)bso_11] ≥ 0)

For Pair COND_1253_0_MAIN_LOAD(TRUE, x1, x2, x0) → 1253_0_MAIN_LOAD(+(x1, -1), x2, +(x0, -1)) the following chains were created:
• We consider the chain COND_1253_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1]) → 1253_0_MAIN_LOAD(+(x1[1], -1), x2[1], +(x0[1], -1)) which results in the following constraint:

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

(11)    ((UIncreasing(1253_0_MAIN_LOAD(+(x1[1], -1), x2[1], +(x0[1], -1))), ≥)∧[(-1)bso_13] ≥ 0)

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

(12)    ((UIncreasing(1253_0_MAIN_LOAD(+(x1[1], -1), x2[1], +(x0[1], -1))), ≥)∧[(-1)bso_13] ≥ 0)

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

(13)    ((UIncreasing(1253_0_MAIN_LOAD(+(x1[1], -1), x2[1], +(x0[1], -1))), ≥)∧[(-1)bso_13] ≥ 0)

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

(14)    ((UIncreasing(1253_0_MAIN_LOAD(+(x1[1], -1), x2[1], +(x0[1], -1))), ≥)∧0 = 0∧0 = 0∧0 = 0∧[(-1)bso_13] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• 1253_0_MAIN_LOAD(x1, x2, x0) → COND_1253_0_MAIN_LOAD(&&(<(x2, x1), <(x2, x0)), x1, x2, x0)
• (x1[0] ≥ 0∧x2[0] ≥ 0∧x0[0] ≥ 0 ⇒ (UIncreasing(COND_1253_0_MAIN_LOAD(&&(<(x2[0], x1[0]), <(x2[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(-1)Bound*bni_10 + bni_10] + [bni_10]x1[0] ≥ 0∧[1 + (-1)bso_11] ≥ 0)
• (x1[0] ≥ 0∧x2[0] ≥ 0∧x0[0] ≥ 0 ⇒ (UIncreasing(COND_1253_0_MAIN_LOAD(&&(<(x2[0], x1[0]), <(x2[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(-1)Bound*bni_10 + bni_10] + [bni_10]x1[0] ≥ 0∧[1 + (-1)bso_11] ≥ 0)

• ((UIncreasing(1253_0_MAIN_LOAD(+(x1[1], -1), x2[1], +(x0[1], -1))), ≥)∧0 = 0∧0 = 0∧0 = 0∧[(-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(1253_0_MAIN_LOAD(x1, x2, x3)) = [-1]x2 + x1
POL(COND_1253_0_MAIN_LOAD(x1, x2, x3, x4)) = [-1] + [-1]x3 + x2
POL(&&(x1, x2)) = [-1]
POL(<(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
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.

### (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:
(1): COND_1253_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1]) → 1253_0_MAIN_LOAD(x1[1] + -1, x2[1], x0[1] + -1)

The set Q is empty.

### (7) IDependencyGraphProof (EQUIVALENT transformation)

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