### (0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: PastaC7
`/** * 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 PastaC7 {    public static void main(String[] args) {        Random.args = args;        int i = Random.random();        int j = Random.random();        int k = Random.random();        while (i <= 100 && j <= k) {            int t = i;            i = j;            j = i + 1;            k--;        }    }}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:
Graph of 259 nodes with 1 SCC.

### (3) FIGtoITRSProof (SOUND transformation)

Transformed FIGraph to ITRS rules

### (4) Obligation:

ITRS 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 TRS R consists of the following rules:
Load1106(i14, i23, i51) → Cond_Load1106(i23 <= i51 && i14 <= 100, i14, i23, i51)
Cond_Load1106(TRUE, i14, i23, i51) → Load1106(i23, i23 + 1, i51 + -1)
The set Q consists of the following terms:

### (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

The ITRS R consists of the following rules:
Load1106(i14, i23, i51) → Cond_Load1106(i23 <= i51 && i14 <= 100, i14, i23, i51)
Cond_Load1106(TRUE, i14, i23, i51) → Load1106(i23, i23 + 1, i51 + -1)

The integer pair graph contains the following rules and edges:
(0): LOAD1106(i14[0], i23[0], i51[0]) → COND_LOAD1106(i23[0] <= i51[0] && i14[0] <= 100, i14[0], i23[0], i51[0])
(1): COND_LOAD1106(TRUE, i14[1], i23[1], i51[1]) → LOAD1106(i23[1], i23[1] + 1, i51[1] + -1)

(0) -> (1), if ((i14[0]* i14[1])∧(i23[0] <= i51[0] && i14[0] <= 100* TRUE)∧(i51[0]* i51[1])∧(i23[0]* i23[1]))

(1) -> (0), if ((i23[1]* i14[0])∧(i51[1] + -1* i51[0])∧(i23[1] + 1* i23[0]))

The set Q consists of the following terms:

### (7) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

### (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): LOAD1106(i14[0], i23[0], i51[0]) → COND_LOAD1106(i23[0] <= i51[0] && i14[0] <= 100, i14[0], i23[0], i51[0])
(1): COND_LOAD1106(TRUE, i14[1], i23[1], i51[1]) → LOAD1106(i23[1], i23[1] + 1, i51[1] + -1)

(0) -> (1), if ((i14[0]* i14[1])∧(i23[0] <= i51[0] && i14[0] <= 100* TRUE)∧(i51[0]* i51[1])∧(i23[0]* i23[1]))

(1) -> (0), if ((i23[1]* i14[0])∧(i51[1] + -1* i51[0])∧(i23[1] + 1* i23[0]))

The set Q consists of the following terms:

### (9) 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 LOAD1106(i14, i23, i51) → COND_LOAD1106(&&(<=(i23, i51), <=(i14, 100)), i14, i23, i51) the following chains were created:
• We consider the chain LOAD1106(i14[0], i23[0], i51[0]) → COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0]), COND_LOAD1106(TRUE, i14[1], i23[1], i51[1]) → LOAD1106(i23[1], +(i23[1], 1), +(i51[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)    (i51[0] + [-1]i23[0] ≥ 0∧[100] + [-1]i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] + [(-1)bni_10]i23[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

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

(4)    (i51[0] + [-1]i23[0] ≥ 0∧[100] + [-1]i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] + [(-1)bni_10]i23[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

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

(5)    (i51[0] + [-1]i23[0] ≥ 0∧[100] + [-1]i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] + [(-1)bni_10]i23[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

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

(6)    (i51[0] ≥ 0∧[100] + [-1]i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

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

(7)    (i51[0] ≥ 0∧[100] + [-1]i14[0] ≥ 0∧i23[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

(8)    (i51[0] ≥ 0∧[100] + [-1]i14[0] ≥ 0∧i23[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

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

(9)    (i51[0] ≥ 0∧[100] + [-1]i14[0] ≥ 0∧i23[0] ≥ 0∧i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

(10)    (i51[0] ≥ 0∧[100] + i14[0] ≥ 0∧i23[0] ≥ 0∧i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

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

(11)    (i51[0] ≥ 0∧[100] + i14[0] ≥ 0∧i23[0] ≥ 0∧i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

(12)    (i51[0] ≥ 0∧[100] + [-1]i14[0] ≥ 0∧i23[0] ≥ 0∧i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

For Pair COND_LOAD1106(TRUE, i14, i23, i51) → LOAD1106(i23, +(i23, 1), +(i51, -1)) the following chains were created:
• We consider the chain COND_LOAD1106(TRUE, i14[1], i23[1], i51[1]) → LOAD1106(i23[1], +(i23[1], 1), +(i51[1], -1)) which results in the following constraint:

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

(14)    ((UIncreasing(LOAD1106(i23[1], +(i23[1], 1), +(i51[1], -1))), ≥)∧[(-1)bso_13] ≥ 0)

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

(15)    ((UIncreasing(LOAD1106(i23[1], +(i23[1], 1), +(i51[1], -1))), ≥)∧[(-1)bso_13] ≥ 0)

We simplified constraint (15) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:

(16)    ((UIncreasing(LOAD1106(i23[1], +(i23[1], 1), +(i51[1], -1))), ≥)∧[(-1)bso_13] ≥ 0)

We simplified constraint (16) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:

(17)    ((UIncreasing(LOAD1106(i23[1], +(i23[1], 1), +(i51[1], -1))), ≥)∧0 = 0∧0 = 0∧0 = 0∧[(-1)bso_13] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD1106(i14, i23, i51) → COND_LOAD1106(&&(<=(i23, i51), <=(i14, 100)), i14, i23, i51)
• (i51[0] ≥ 0∧[100] + [-1]i14[0] ≥ 0∧i23[0] ≥ 0∧i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)
• (i51[0] ≥ 0∧[100] + i14[0] ≥ 0∧i23[0] ≥ 0∧i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)
• (i51[0] ≥ 0∧[100] + i14[0] ≥ 0∧i23[0] ≥ 0∧i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)
• (i51[0] ≥ 0∧[100] + [-1]i14[0] ≥ 0∧i23[0] ≥ 0∧i14[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1106(&&(<=(i23[0], i51[0]), <=(i14[0], 100)), i14[0], i23[0], i51[0])), ≥)∧[bni_10 + (-1)Bound*bni_10] + [bni_10]i51[0] ≥ 0∧[2 + (-1)bso_11] ≥ 0)

• ((UIncreasing(LOAD1106(i23[1], +(i23[1], 1), +(i51[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(LOAD1106(x1, x2, x3)) = [1] + x3 + [-1]x2
POL(COND_LOAD1106(x1, x2, x3, x4)) = [-1] + x4 + [-1]x3
POL(&&(x1, x2)) = [-1]
POL(<=(x1, x2)) = [-1]
POL(100) = [100]
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.

### (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:
(1): COND_LOAD1106(TRUE, i14[1], i23[1], i51[1]) → LOAD1106(i23[1], i23[1] + 1, i51[1] + -1)

The set Q consists of the following terms: