Termination proof of /tmp/tmp1XpXQM/PastaA7.jar

Termination of the given JBC could be proven:

0 JBC

↳1 JBC2FIG (⇒)

↳2 JBCTerminationGraph

↳3 FIGtoITRSProof (⇒)

↳4 IDP

↳5 IDPNonInfProof (⇒)

↳6 AND

    ↳7 IDP

        ↳8 IDependencyGraphProof (⇔)

        ↳9 TRUE

    ↳10 IDP

        ↳11 IDependencyGraphProof (⇔)

        ↳12 TRUE

(0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: PastaA7

/**
 * 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 PastaA7 {
    public static void main(String[] args) {
        Random.args = args;
        int x = Random.random();
        int y = Random.random();
        int z = Random.random();

        while (x > y && x > z) {
            y++;
            z++;
        }
    }
}


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:
PastaA7.main([Ljava/lang/String;)V: Graph of 227 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:

1023_0_main_Load(x1, x2, x3, x4, x5) → 1023_0_main_Load(x2, x3, x4, x5)
Cond_1023_0_main_Load(x1, x2, x3, x4, x5, x6) → Cond_1023_0_main_Load(x1, x3, x4, x5, x6)

Filtered duplicate args:

1023_0_main_Load(x1, x2, x3, x4) → 1023_0_main_Load(x2, x3, x4)
Cond_1023_0_main_Load(x1, x2, x3, x4, x5) → Cond_1023_0_main_Load(x1, x3, x4, x5)

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): 1023_0_MAIN_LOAD(x1[0], x2[0], x0[0]) → COND_1023_0_MAIN_LOAD(x2[0] >= 0 && x2[0] < x0[0] && x1[0] >= 0 && x1[0] < x0[0], x1[0], x2[0], x0[0])
(1): COND_1023_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1]) → 1023_0_MAIN_LOAD(x1[1] + 1, x2[1] + 1, x0[1])

(0) -> (1), if ((x2[0] >= 0 && x2[0] < x0[0] && x1[0] >= 0 && x1[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] + 1 →^* x2[0])∧(x0[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 1023_0_MAIN_LOAD(x1, x2, x0) → COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2, 0), <(x2, x0)), >=(x1, 0)), <(x1, x0)), x1, x2, x0) the following chains were created:

We consider the chain 1023_0_MAIN_LOAD(x1[0], x2[0], x0[0]) → COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0]), COND_1023_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1]) → 1023_0_MAIN_LOAD(+(x1[1], 1), +(x2[1], 1), x0[1]) which results in the following constraint:
(1)    (&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0]))=TRUE∧x1[0]=x1[1]∧x2[0]=x2[1]∧x0[0]=x0[1] ⇒ 1023_0_MAIN_LOAD(x1[0], x2[0], x0[0])≥NonInfC∧1023_0_MAIN_LOAD(x1[0], x2[0], x0[0])≥COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])∧(U^Increasing(COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])), ≥))

We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint:
(2)    (<(x1[0], x0[0])=TRUE∧>=(x1[0], 0)=TRUE∧>=(x2[0], 0)=TRUE∧<(x2[0], x0[0])=TRUE ⇒ 1023_0_MAIN_LOAD(x1[0], x2[0], x0[0])≥NonInfC∧1023_0_MAIN_LOAD(x1[0], x2[0], x0[0])≥COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])∧(U^Increasing(COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])), ≥))

We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(3)    (x0[0] + [-1] + [-1]x1[0] ≥ 0∧x1[0] ≥ 0∧x2[0] ≥ 0∧x0[0] + [-1] + [-1]x2[0] ≥ 0 ⇒ (U^Increasing(COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[bni_11 + (-1)Bound*bni_11] + [(2)bni_11]x0[0] + [(-1)bni_11]x2[0] + [(-1)bni_11]x1[0] ≥ 0∧[(-1)bso_12] ≥ 0)

We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(4)    (x0[0] + [-1] + [-1]x1[0] ≥ 0∧x1[0] ≥ 0∧x2[0] ≥ 0∧x0[0] + [-1] + [-1]x2[0] ≥ 0 ⇒ (U^Increasing(COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[bni_11 + (-1)Bound*bni_11] + [(2)bni_11]x0[0] + [(-1)bni_11]x2[0] + [(-1)bni_11]x1[0] ≥ 0∧[(-1)bso_12] ≥ 0)

We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5)    (x0[0] + [-1] + [-1]x1[0] ≥ 0∧x1[0] ≥ 0∧x2[0] ≥ 0∧x0[0] + [-1] + [-1]x2[0] ≥ 0 ⇒ (U^Increasing(COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[bni_11 + (-1)Bound*bni_11] + [(2)bni_11]x0[0] + [(-1)bni_11]x2[0] + [(-1)bni_11]x1[0] ≥ 0∧[(-1)bso_12] ≥ 0)

We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(6)    (x0[0] ≥ 0∧x1[0] ≥ 0∧x2[0] ≥ 0∧x1[0] + x0[0] + [-1]x2[0] ≥ 0 ⇒ (U^Increasing(COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(3)bni_11 + (-1)Bound*bni_11] + [bni_11]x1[0] + [(2)bni_11]x0[0] + [(-1)bni_11]x2[0] ≥ 0∧[(-1)bso_12] ≥ 0)

For Pair COND_1023_0_MAIN_LOAD(TRUE, x1, x2, x0) → 1023_0_MAIN_LOAD(+(x1, 1), +(x2, 1), x0) the following chains were created:

We consider the chain COND_1023_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1]) → 1023_0_MAIN_LOAD(+(x1[1], 1), +(x2[1], 1), x0[1]) which results in the following constraint:
(7)    (COND_1023_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1])≥NonInfC∧COND_1023_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1])≥1023_0_MAIN_LOAD(+(x1[1], 1), +(x2[1], 1), x0[1])∧(U^Increasing(1023_0_MAIN_LOAD(+(x1[1], 1), +(x2[1], 1), x0[1])), ≥))

We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(8)    ((U^Increasing(1023_0_MAIN_LOAD(+(x1[1], 1), +(x2[1], 1), x0[1])), ≥)∧[2 + (-1)bso_14] ≥ 0)

We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(9)    ((U^Increasing(1023_0_MAIN_LOAD(+(x1[1], 1), +(x2[1], 1), x0[1])), ≥)∧[2 + (-1)bso_14] ≥ 0)

We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(10)    ((U^Increasing(1023_0_MAIN_LOAD(+(x1[1], 1), +(x2[1], 1), x0[1])), ≥)∧[2 + (-1)bso_14] ≥ 0)

We simplified constraint (10) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(11)    ((U^Increasing(1023_0_MAIN_LOAD(+(x1[1], 1), +(x2[1], 1), x0[1])), ≥)∧0 = 0∧0 = 0∧0 = 0∧[2 + (-1)bso_14] ≥ 0)

To summarize, we get the following constraints P_≥ for the following pairs.

1023_0_MAIN_LOAD(x1, x2, x0) → COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2, 0), <(x2, x0)), >=(x1, 0)), <(x1, x0)), x1, x2, x0)
- (x0[0] ≥ 0∧x1[0] ≥ 0∧x2[0] ≥ 0∧x1[0] + x0[0] + [-1]x2[0] ≥ 0 ⇒ (U^Increasing(COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])), ≥)∧[(3)bni_11 + (-1)Bound*bni_11] + [bni_11]x1[0] + [(2)bni_11]x0[0] + [(-1)bni_11]x2[0] ≥ 0∧[(-1)bso_12] ≥ 0)
COND_1023_0_MAIN_LOAD(TRUE, x1, x2, x0) → 1023_0_MAIN_LOAD(+(x1, 1), +(x2, 1), x0)
- ((U^Increasing(1023_0_MAIN_LOAD(+(x1[1], 1), +(x2[1], 1), x0[1])), ≥)∧0 = 0∧0 = 0∧0 = 0∧[2 + (-1)bso_14] ≥ 0)

The constraints for P_> respective P_bound 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 P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:

POL(TRUE) = 0
POL(FALSE) = 0
POL(1023_0_MAIN_LOAD(x₁, x₂, x₃)) = [1] + [2]x₃ + [-1]x₂ + [-1]x₁
POL(COND_1023_0_MAIN_LOAD(x₁, x₂, x₃, x₄)) = [1] + [2]x₄ + [-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(1) = [1]

The following pairs are in P_>:

COND_1023_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1]) → 1023_0_MAIN_LOAD(+(x1[1], 1), +(x2[1], 1), x0[1])

The following pairs are in P_bound:

1023_0_MAIN_LOAD(x1[0], x2[0], x0[0]) → COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])

The following pairs are in P_≥:

1023_0_MAIN_LOAD(x1[0], x2[0], x0[0]) → COND_1023_0_MAIN_LOAD(&&(&&(&&(>=(x2[0], 0), <(x2[0], x0[0])), >=(x1[0], 0)), <(x1[0], x0[0])), x1[0], x2[0], x0[0])

There are no usable rules.

(6) Complex Obligation (AND)

(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

(8) IDependencyGraphProof (EQUIVALENT transformation)

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

(9) TRUE

(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_1023_0_MAIN_LOAD(TRUE, x1[1], x2[1], x0[1]) → 1023_0_MAIN_LOAD(x1[1] + 1, x2[1] + 1, x0[1])

The set Q is empty.

(11) IDependencyGraphProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) JBC2FIG (SOUND transformation)

(2) Obligation:

(3) FIGtoITRSProof (SOUND transformation)

(4) Obligation:

(5) IDPNonInfProof (SOUND transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) IDependencyGraphProof (EQUIVALENT transformation)

(9) TRUE

(10) Obligation:

(11) IDependencyGraphProof (EQUIVALENT transformation)

(12) TRUE