### (0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: PastaB1
`/** * 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 PastaB1 {    public static void main(String[] args) {        Random.args = args;        int x = Random.random();        int y = Random.random();        while (x > y) {            x--;        }    }}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 184 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:
Load599(i11, i26) → Cond_Load599(i11 > i26, i11, i26)
Cond_Load599(TRUE, i11, i26) → Load599(i11 + -1, i26)
The set Q consists of the following terms:
Load599(x0, x1)
Cond_Load599(TRUE, x0, x1)

### (5) ITRStoIDPProof (EQUIVALENT transformation)

Added dependency pairs

### (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

The ITRS R consists of the following rules:
Load599(i11, i26) → Cond_Load599(i11 > i26, i11, i26)
Cond_Load599(TRUE, i11, i26) → Load599(i11 + -1, i26)

The integer pair graph contains the following rules and edges:
(0): LOAD599(i11[0], i26[0]) → COND_LOAD599(i11[0] > i26[0], i11[0], i26[0])
(1): COND_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(i11[1] + -1, i26[1])

(0) -> (1), if ((i11[0] > i26[0]* TRUE)∧(i11[0]* i11[1])∧(i26[0]* i26[1]))

(1) -> (0), if ((i11[1] + -1* i11[0])∧(i26[1]* i26[0]))

The set Q consists of the following terms:
Load599(x0, x1)
Cond_Load599(TRUE, x0, x1)

### (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:

Integer

R is empty.

The integer pair graph contains the following rules and edges:
(0): LOAD599(i11[0], i26[0]) → COND_LOAD599(i11[0] > i26[0], i11[0], i26[0])
(1): COND_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(i11[1] + -1, i26[1])

(0) -> (1), if ((i11[0] > i26[0]* TRUE)∧(i11[0]* i11[1])∧(i26[0]* i26[1]))

(1) -> (0), if ((i11[1] + -1* i11[0])∧(i26[1]* i26[0]))

The set Q consists of the following terms:
Load599(x0, x1)
Cond_Load599(TRUE, x0, x1)

### (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 LOAD599(i11, i26) → COND_LOAD599(>(i11, i26), i11, i26) the following chains were created:
• We consider the chain LOAD599(i11[0], i26[0]) → COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0]), COND_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(+(i11[1], -1), i26[1]) which results in the following constraint:

(1)    (>(i11[0], i26[0])=TRUEi11[0]=i11[1]i26[0]=i26[1]LOAD599(i11[0], i26[0])≥NonInfC∧LOAD599(i11[0], i26[0])≥COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])∧(UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥))

We simplified constraint (1) using rule (IV) which results in the following new constraint:

(2)    (>(i11[0], i26[0])=TRUELOAD599(i11[0], i26[0])≥NonInfC∧LOAD599(i11[0], i26[0])≥COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])∧(UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥))

We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:

(3)    (i11[0] + [-1] + [-1]i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[bni_8 + (-1)Bound*bni_8] + [(-1)bni_8]i26[0] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)

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

(4)    (i11[0] + [-1] + [-1]i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[bni_8 + (-1)Bound*bni_8] + [(-1)bni_8]i26[0] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)

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

(5)    (i11[0] + [-1] + [-1]i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[bni_8 + (-1)Bound*bni_8] + [(-1)bni_8]i26[0] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)

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

(6)    (i11[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)

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

(7)    (i11[0] ≥ 0∧i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)

(8)    (i11[0] ≥ 0∧i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)

For Pair COND_LOAD599(TRUE, i11, i26) → LOAD599(+(i11, -1), i26) the following chains were created:
• We consider the chain COND_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(+(i11[1], -1), i26[1]) which results in the following constraint:

(9)    (COND_LOAD599(TRUE, i11[1], i26[1])≥NonInfC∧COND_LOAD599(TRUE, i11[1], i26[1])≥LOAD599(+(i11[1], -1), i26[1])∧(UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥))

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

(10)    ((UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥)∧[(-1)bso_11] ≥ 0)

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

(11)    ((UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥)∧[(-1)bso_11] ≥ 0)

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

(12)    ((UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥)∧[(-1)bso_11] ≥ 0)

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

(13)    ((UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥)∧0 = 0∧0 = 0∧[(-1)bso_11] ≥ 0)

To summarize, we get the following constraints P for the following pairs.
• LOAD599(i11, i26) → COND_LOAD599(>(i11, i26), i11, i26)
• (i11[0] ≥ 0∧i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)
• (i11[0] ≥ 0∧i26[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])), ≥)∧[(2)bni_8 + (-1)Bound*bni_8] + [bni_8]i11[0] ≥ 0∧[1 + (-1)bso_9] ≥ 0)

• COND_LOAD599(TRUE, i11, i26) → LOAD599(+(i11, -1), i26)
• ((UIncreasing(LOAD599(+(i11[1], -1), i26[1])), ≥)∧0 = 0∧0 = 0∧[(-1)bso_11] ≥ 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(LOAD599(x1, x2)) = [1] + [-1]x2 + x1
POL(COND_LOAD599(x1, x2, x3)) = [-1]x3 + x2
POL(>(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(-1) = [-1]

The following pairs are in P>:

LOAD599(i11[0], i26[0]) → COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])

The following pairs are in Pbound:

LOAD599(i11[0], i26[0]) → COND_LOAD599(>(i11[0], i26[0]), i11[0], i26[0])

The following pairs are in P:

COND_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(+(i11[1], -1), i26[1])

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_LOAD599(TRUE, i11[1], i26[1]) → LOAD599(i11[1] + -1, i26[1])

The set Q consists of the following terms:
Load599(x0, x1)
Cond_Load599(TRUE, x0, x1)

### (11) IDependencyGraphProof (EQUIVALENT transformation)

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