### (0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: Duplicate
`public class Duplicate{  public static int round (int x) {    if (x % 2 == 0) return x;    else return x+1;  }  public static void main(String[] args) {    Random.args = args;    int x = Random.random();    int y = Random.random();    while ((x > y) && (y > 2)) {      x++;      y = 2*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:
Graph of 194 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:
Load551(2, i11, i21) → Cond_Load551(i21 > 2 && i11 > i21, 2, i11, i21)
Cond_Load551(TRUE, 2, i11, i21) → Load551(2, i11 + 1, 2 * i21)
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:
Load551(2, i11, i21) → Cond_Load551(i21 > 2 && i11 > i21, 2, i11, i21)
Cond_Load551(TRUE, 2, i11, i21) → Load551(2, i11 + 1, 2 * i21)

The integer pair graph contains the following rules and edges:
(0): LOAD551(2, i11[0], i21[0]) → COND_LOAD551(i21[0] > 2 && i11[0] > i21[0], 2, i11[0], i21[0])
(1): COND_LOAD551(TRUE, 2, i11[1], i21[1]) → LOAD551(2, i11[1] + 1, 2 * i21[1])

(0) -> (1), if ((i11[0]* i11[1])∧(i21[0] > 2 && i11[0] > i21[0]* TRUE)∧(i21[0]* i21[1]))

(1) -> (0), if ((2 * i21[1]* i21[0])∧(i11[1] + 1* i11[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): LOAD551(2, i11[0], i21[0]) → COND_LOAD551(i21[0] > 2 && i11[0] > i21[0], 2, i11[0], i21[0])
(1): COND_LOAD551(TRUE, 2, i11[1], i21[1]) → LOAD551(2, i11[1] + 1, 2 * i21[1])

(0) -> (1), if ((i11[0]* i11[1])∧(i21[0] > 2 && i11[0] > i21[0]* TRUE)∧(i21[0]* i21[1]))

(1) -> (0), if ((2 * i21[1]* i21[0])∧(i11[1] + 1* i11[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 LOAD551(2, i11, i21) → COND_LOAD551(&&(>(i21, 2), >(i11, i21)), 2, i11, i21) the following chains were created:
• We consider the chain LOAD551(2, i11[0], i21[0]) → COND_LOAD551(&&(>(i21[0], 2), >(i11[0], i21[0])), 2, i11[0], i21[0]), COND_LOAD551(TRUE, 2, i11[1], i21[1]) → LOAD551(2, +(i11[1], 1), *(2, i21[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)    (i21[0] + [-3] ≥ 0∧i11[0] + [-1] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD551(&&(>(i21[0], 2), >(i11[0], i21[0])), 2, i11[0], i21[0])), ≥)∧[bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i21[0] + [bni_14]i11[0] ≥ 0∧[1 + (-1)bso_15] ≥ 0)

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

(4)    (i21[0] + [-3] ≥ 0∧i11[0] + [-1] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD551(&&(>(i21[0], 2), >(i11[0], i21[0])), 2, i11[0], i21[0])), ≥)∧[bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i21[0] + [bni_14]i11[0] ≥ 0∧[1 + (-1)bso_15] ≥ 0)

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

(5)    (i21[0] + [-3] ≥ 0∧i11[0] + [-1] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD551(&&(>(i21[0], 2), >(i11[0], i21[0])), 2, i11[0], i21[0])), ≥)∧[bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i21[0] + [bni_14]i11[0] ≥ 0∧[1 + (-1)bso_15] ≥ 0)

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

(6)    (i21[0] ≥ 0∧i11[0] + [-4] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD551(&&(>(i21[0], 2), >(i11[0], i21[0])), 2, i11[0], i21[0])), ≥)∧[(-2)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i21[0] + [bni_14]i11[0] ≥ 0∧[1 + (-1)bso_15] ≥ 0)

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

(7)    (i21[0] ≥ 0∧i11[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD551(&&(>(i21[0], 2), >(i11[0], i21[0])), 2, i11[0], i21[0])), ≥)∧[(2)bni_14 + (-1)Bound*bni_14] + [bni_14]i11[0] ≥ 0∧[1 + (-1)bso_15] ≥ 0)

For Pair COND_LOAD551(TRUE, 2, i11, i21) → LOAD551(2, +(i11, 1), *(2, i21)) the following chains were created:
• We consider the chain LOAD551(2, i11[0], i21[0]) → COND_LOAD551(&&(>(i21[0], 2), >(i11[0], i21[0])), 2, i11[0], i21[0]), COND_LOAD551(TRUE, 2, i11[1], i21[1]) → LOAD551(2, +(i11[1], 1), *(2, i21[1])), LOAD551(2, i11[0], i21[0]) → COND_LOAD551(&&(>(i21[0], 2), >(i11[0], i21[0])), 2, i11[0], i21[0]) which results in the following constraint:

We simplified constraint (8) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:

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

(10)    (i21[0] + [-3] ≥ 0∧i11[0] + [-1] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(LOAD551(2, +(i11[1], 1), *(2, i21[1]))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i21[0] + [bni_16]i11[0] ≥ 0∧[-2 + (-1)bso_17] + i21[0] ≥ 0)

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

(11)    (i21[0] + [-3] ≥ 0∧i11[0] + [-1] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(LOAD551(2, +(i11[1], 1), *(2, i21[1]))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i21[0] + [bni_16]i11[0] ≥ 0∧[-2 + (-1)bso_17] + i21[0] ≥ 0)

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

(12)    (i21[0] + [-3] ≥ 0∧i11[0] + [-1] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(LOAD551(2, +(i11[1], 1), *(2, i21[1]))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i21[0] + [bni_16]i11[0] ≥ 0∧[-2 + (-1)bso_17] + i21[0] ≥ 0)

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

(13)    (i21[0] ≥ 0∧i11[0] + [-4] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(LOAD551(2, +(i11[1], 1), *(2, i21[1]))), ≥)∧[(-1)Bound*bni_16 + (-3)bni_16] + [(-1)bni_16]i21[0] + [bni_16]i11[0] ≥ 0∧[1 + (-1)bso_17] + i21[0] ≥ 0)

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

(14)    (i21[0] ≥ 0∧i11[0] ≥ 0 ⇒ (UIncreasing(LOAD551(2, +(i11[1], 1), *(2, i21[1]))), ≥)∧[(-1)Bound*bni_16 + bni_16] + [bni_16]i11[0] ≥ 0∧[1 + (-1)bso_17] + i21[0] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD551(2, i11, i21) → COND_LOAD551(&&(>(i21, 2), >(i11, i21)), 2, i11, i21)
• (i21[0] ≥ 0∧i11[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD551(&&(>(i21[0], 2), >(i11[0], i21[0])), 2, i11[0], i21[0])), ≥)∧[(2)bni_14 + (-1)Bound*bni_14] + [bni_14]i11[0] ≥ 0∧[1 + (-1)bso_15] ≥ 0)

• (i21[0] ≥ 0∧i11[0] ≥ 0 ⇒ (UIncreasing(LOAD551(2, +(i11[1], 1), *(2, i21[1]))), ≥)∧[(-1)Bound*bni_16 + bni_16] + [bni_16]i11[0] ≥ 0∧[1 + (-1)bso_17] + i21[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) = [2]
POL(FALSE) = 0
POL(LOAD551(x1, x2, x3)) = [-1] + [-1]x3 + x2 + x1
POL(2) = [2]
POL(COND_LOAD551(x1, x2, x3, x4)) = [-1]x4 + x3
POL(&&(x1, x2)) = [-1]
POL(>(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(1) = [1]
POL(*(x1, x2)) = x1·x2

The following pairs are in P>:

The following pairs are in Pbound:

The following pairs are in P:
none

At least the following rules have been oriented under context sensitive arithmetic replacement:

FALSE1&&(FALSE, TRUE)1

### (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:
none

R is empty.

The integer pair graph is empty.

The set Q consists of the following terms: