(0) Obligation:

JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_22 (Sun Microsystems Inc.) Main-Class: LeUserDefRec
public class LeUserDefRec {
public static void main(String[] args) {
int x = args[0].length();
int y = args[1].length();
le(x, y);
}

public static boolean le(int x, int y) {
if (x > 0 && y > 0) {
return le(x-1, y-1);
} else {
return (x == 0);
}
}
}


(1) JBC2FIG (SOUND transformation)

Constructed FIGraph.

(2) Obligation:

FIGraph based on JBC Program:
LeUserDefRec.main([Ljava/lang/String;)V: Graph of 129 nodes with 0 SCCs.

LeUserDefRec.le(II)Z: Graph of 35 nodes with 0 SCCs.


(3) FIGtoITRSProof (SOUND transformation)

Transformed FIGraph SCCs to IDPs. Logs:


Log for SCC 0:

Generated 15 rules for P and 19 rules for R.


Combined rules. Obtained 1 rules for P and 4 rules for R.


Filtered ground terms:


415_0_le_LE(x1, x2, x3, x4) → 415_0_le_LE(x2, x3, x4)
Cond_415_0_le_LE(x1, x2, x3, x4, x5) → Cond_415_0_le_LE(x1, x3, x4, x5)
689_0_le_Return(x1, x2) → 689_0_le_Return(x2)
643_0_le_Return(x1, x2) → 643_0_le_Return
616_0_le_Return(x1) → 616_0_le_Return

Filtered duplicate args:


415_0_le_LE(x1, x2, x3) → 415_0_le_LE(x2, x3)
Cond_415_0_le_LE(x1, x2, x3, x4) → Cond_415_0_le_LE(x1, x3, x4)

Filtered unneeded arguments:


681_1_le_InvokeMethod(x1, x2, x3, x4) → 681_1_le_InvokeMethod(x1, x3, x4)

Combined rules. Obtained 1 rules for P and 4 rules for R.


Finished conversion. Obtained 1 rules for P and 4 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


The ITRS R consists of the following rules:
415_0_le_LE(x1, 0) → 616_0_le_Return
681_1_le_InvokeMethod(616_0_le_Return, 0, x2) → 689_0_le_Return(x0)
681_1_le_InvokeMethod(643_0_le_Return, x2, 0) → 689_0_le_Return(x1)
681_1_le_InvokeMethod(689_0_le_Return(x0), x0, x2) → 689_0_le_Return(x1)

The integer pair graph contains the following rules and edges:
(0): 415_0_LE_LE(x1[0], x0[0]) → COND_415_0_LE_LE(x1[0] > 0 && x0[0] > 0, x1[0], x0[0])
(1): COND_415_0_LE_LE(TRUE, x1[1], x0[1]) → 415_0_LE_LE(x1[1] - 1, x0[1] - 1)

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


(1) -> (0), if ((x1[1] - 1* x1[0])∧(x0[1] - 1* x0[0]))



The set Q consists of the following terms:
415_0_le_LE(x0, 0)
681_1_le_InvokeMethod(616_0_le_Return, 0, x0)
681_1_le_InvokeMethod(643_0_le_Return, x0, 0)
681_1_le_InvokeMethod(689_0_le_Return(x0), x0, x1)

(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 415_0_LE_LE(x1, x0) → COND_415_0_LE_LE(&&(>(x1, 0), >(x0, 0)), x1, x0) the following chains were created:
  • We consider the chain 415_0_LE_LE(x1[0], x0[0]) → COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0]), COND_415_0_LE_LE(TRUE, x1[1], x0[1]) → 415_0_LE_LE(-(x1[1], 1), -(x0[1], 1)) which results in the following constraint:

    (1)    (&&(>(x1[0], 0), >(x0[0], 0))=TRUEx1[0]=x1[1]x0[0]=x0[1]415_0_LE_LE(x1[0], x0[0])≥NonInfC∧415_0_LE_LE(x1[0], x0[0])≥COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])∧(UIncreasing(COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])), ≥))



    We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint:

    (2)    (>(x1[0], 0)=TRUE>(x0[0], 0)=TRUE415_0_LE_LE(x1[0], x0[0])≥NonInfC∧415_0_LE_LE(x1[0], x0[0])≥COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])∧(UIncreasing(COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])), ≥))



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

    (3)    (x1[0] + [-1] ≥ 0∧x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])), ≥)∧[bni_14 + (-1)Bound*bni_14] + [(2)bni_14]x0[0] + [(2)bni_14]x1[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (4)    (x1[0] + [-1] ≥ 0∧x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])), ≥)∧[bni_14 + (-1)Bound*bni_14] + [(2)bni_14]x0[0] + [(2)bni_14]x1[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (5)    (x1[0] + [-1] ≥ 0∧x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])), ≥)∧[bni_14 + (-1)Bound*bni_14] + [(2)bni_14]x0[0] + [(2)bni_14]x1[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (6)    (x1[0] ≥ 0∧x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])), ≥)∧[(3)bni_14 + (-1)Bound*bni_14] + [(2)bni_14]x0[0] + [(2)bni_14]x1[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (7)    (x1[0] ≥ 0∧x0[0] ≥ 0 ⇒ (UIncreasing(COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])), ≥)∧[(5)bni_14 + (-1)Bound*bni_14] + [(2)bni_14]x0[0] + [(2)bni_14]x1[0] ≥ 0∧[(-1)bso_15] ≥ 0)







For Pair COND_415_0_LE_LE(TRUE, x1, x0) → 415_0_LE_LE(-(x1, 1), -(x0, 1)) the following chains were created:
  • We consider the chain COND_415_0_LE_LE(TRUE, x1[1], x0[1]) → 415_0_LE_LE(-(x1[1], 1), -(x0[1], 1)) which results in the following constraint:

    (8)    (COND_415_0_LE_LE(TRUE, x1[1], x0[1])≥NonInfC∧COND_415_0_LE_LE(TRUE, x1[1], x0[1])≥415_0_LE_LE(-(x1[1], 1), -(x0[1], 1))∧(UIncreasing(415_0_LE_LE(-(x1[1], 1), -(x0[1], 1))), ≥))



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

    (9)    ((UIncreasing(415_0_LE_LE(-(x1[1], 1), -(x0[1], 1))), ≥)∧[4 + (-1)bso_17] ≥ 0)



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

    (10)    ((UIncreasing(415_0_LE_LE(-(x1[1], 1), -(x0[1], 1))), ≥)∧[4 + (-1)bso_17] ≥ 0)



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

    (11)    ((UIncreasing(415_0_LE_LE(-(x1[1], 1), -(x0[1], 1))), ≥)∧[4 + (-1)bso_17] ≥ 0)



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

    (12)    ((UIncreasing(415_0_LE_LE(-(x1[1], 1), -(x0[1], 1))), ≥)∧0 = 0∧0 = 0∧[4 + (-1)bso_17] ≥ 0)







To summarize, we get the following constraints P for the following pairs.
  • 415_0_LE_LE(x1, x0) → COND_415_0_LE_LE(&&(>(x1, 0), >(x0, 0)), x1, x0)
    • (x1[0] ≥ 0∧x0[0] ≥ 0 ⇒ (UIncreasing(COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])), ≥)∧[(5)bni_14 + (-1)Bound*bni_14] + [(2)bni_14]x0[0] + [(2)bni_14]x1[0] ≥ 0∧[(-1)bso_15] ≥ 0)

  • COND_415_0_LE_LE(TRUE, x1, x0) → 415_0_LE_LE(-(x1, 1), -(x0, 1))
    • ((UIncreasing(415_0_LE_LE(-(x1[1], 1), -(x0[1], 1))), ≥)∧0 = 0∧0 = 0∧[4 + (-1)bso_17] ≥ 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(415_0_le_LE(x1, x2)) = [-1]   
POL(0) = 0   
POL(616_0_le_Return) = [-1]   
POL(681_1_le_InvokeMethod(x1, x2, x3)) = [-1]   
POL(689_0_le_Return(x1)) = [-1]   
POL(643_0_le_Return) = [-1]   
POL(415_0_LE_LE(x1, x2)) = [1] + [2]x2 + [2]x1   
POL(COND_415_0_LE_LE(x1, x2, x3)) = [1] + [2]x3 + [2]x2   
POL(&&(x1, x2)) = [-1]   
POL(>(x1, x2)) = [-1]   
POL(-(x1, x2)) = x1 + [-1]x2   
POL(1) = [1]   

The following pairs are in P>:

COND_415_0_LE_LE(TRUE, x1[1], x0[1]) → 415_0_LE_LE(-(x1[1], 1), -(x0[1], 1))

The following pairs are in Pbound:

415_0_LE_LE(x1[0], x0[0]) → COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[0], x0[0])

The following pairs are in P:

415_0_LE_LE(x1[0], x0[0]) → COND_415_0_LE_LE(&&(>(x1[0], 0), >(x0[0], 0)), x1[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


The ITRS R consists of the following rules:
415_0_le_LE(x1, 0) → 616_0_le_Return
681_1_le_InvokeMethod(616_0_le_Return, 0, x2) → 689_0_le_Return(x0)
681_1_le_InvokeMethod(643_0_le_Return, x2, 0) → 689_0_le_Return(x1)
681_1_le_InvokeMethod(689_0_le_Return(x0), x0, x2) → 689_0_le_Return(x1)

The integer pair graph contains the following rules and edges:
(0): 415_0_LE_LE(x1[0], x0[0]) → COND_415_0_LE_LE(x1[0] > 0 && x0[0] > 0, x1[0], x0[0])


The set Q consists of the following terms:
415_0_le_LE(x0, 0)
681_1_le_InvokeMethod(616_0_le_Return, 0, x0)
681_1_le_InvokeMethod(643_0_le_Return, x0, 0)
681_1_le_InvokeMethod(689_0_le_Return(x0), x0, x1)

(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


The ITRS R consists of the following rules:
415_0_le_LE(x1, 0) → 616_0_le_Return
681_1_le_InvokeMethod(616_0_le_Return, 0, x2) → 689_0_le_Return(x0)
681_1_le_InvokeMethod(643_0_le_Return, x2, 0) → 689_0_le_Return(x1)
681_1_le_InvokeMethod(689_0_le_Return(x0), x0, x2) → 689_0_le_Return(x1)

The integer pair graph contains the following rules and edges:
(1): COND_415_0_LE_LE(TRUE, x1[1], x0[1]) → 415_0_LE_LE(x1[1] - 1, x0[1] - 1)


The set Q consists of the following terms:
415_0_le_LE(x0, 0)
681_1_le_InvokeMethod(616_0_le_Return, 0, x0)
681_1_le_InvokeMethod(643_0_le_Return, x0, 0)
681_1_le_InvokeMethod(689_0_le_Return(x0), x0, x1)

(11) IDependencyGraphProof (EQUIVALENT transformation)

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

(12) TRUE