Termination of the given JBC could be proven:

0 JBC
↳1 JBC2FIG (⇐)
↳2 FIGraph
↳3 FIGtoITRSProof (⇐)
↳4 AND
    ↳5 ITRS
        ↳6 GroundTermsRemoverProof (⇔)
        ↳7 ITRS
        ↳8 ITRStoIDPProof (⇔)
        ↳9 IDP
        ↳10 UsableRulesProof (⇔)
        ↳11 IDP
        ↳12 IDPNonInfProof (⇐)
        ↳13 AND
            ↳14 IDP
                ↳15 IDependencyGraphProof (⇔)
                ↳16 TRUE
            ↳17 IDP
                ↳18 IDependencyGraphProof (⇔)
                ↳19 TRUE
    ↳20 ITRS
        ↳21 GroundTermsRemoverProof (⇔)
        ↳22 ITRS
        ↳23 ITRStoIDPProof (⇔)
        ↳24 IDP
        ↳25 UsableRulesProof (⇔)
        ↳26 IDP
        ↳27 IDPNonInfProof (⇐)
        ↳28 AND
            ↳29 IDP
                ↳30 IDependencyGraphProof (⇔)
                ↳31 TRUE
            ↳32 IDP
                ↳33 IDependencyGraphProof (⇔)
                ↳34 TRUE
    ↳35 ITRS
        ↳36 GroundTermsRemoverProof (⇔)
        ↳37 ITRS
        ↳38 ITRStoIDPProof (⇔)
        ↳39 IDP
        ↳40 UsableRulesProof (⇔)
        ↳41 IDP
        ↳42 IDPNonInfProof (⇐)
        ↳43 AND
            ↳44 IDP
                ↳45 IDependencyGraphProof (⇔)
                ↳46 TRUE
            ↳47 IDP
                ↳48 IDependencyGraphProof (⇔)
                ↳49 TRUE

(0) Obligation:

JBC Problem based on JBC Program:
No human-readable program information known.

Manifest-Version: 1.0 Created-By: 1.6.0_16 (Sun Microsystems Inc.) Main-Class: example_1/Test

(1) JBC2FIG (SOUND transformation)

Constructed FIGraph.

(2) Obligation:

FIGraph based on JBC Program:
Graph of 165 nodes with 3 SCCs.

(3) FIGtoITRSProof (SOUND transformation)

Transformed FIGraph to ITRS rules

(4) Complex Obligation (AND)

(5) 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:
Load850(1000, 1000, i47, i48) → Cond_Load850(i48 >= 0 && i48 <= 1000 && i48 + 3 > 0, 1000, 1000, i47, i48)
Cond_Load850(TRUE, 1000, 1000, i47, i48) → Load850(1000, 1000, i47 + i48, i48 + 3)
The set Q consists of the following terms:
Load850(1000, 1000, x0, x1)
Cond_Load850(TRUE, 1000, 1000, x0, x1)

(6) GroundTermsRemoverProof (EQUIVALENT transformation)

Some arguments are removed because they always contain the same ground term.
We removed the following ground terms:
  • 1000

We removed arguments according to the following replacements:

Load850(x1, x2, x3, x4) → Load850(x3, x4)
Cond_Load850(x1, x2, x3, x4, x5) → Cond_Load850(x1, x4, x5)

(7) 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:
Load850(i47, i48) → Cond_Load850(i48 >= 0 && i48 <= 1000 && i48 + 3 > 0, i47, i48)
Cond_Load850(TRUE, i47, i48) → Load850(i47 + i48, i48 + 3)
The set Q consists of the following terms:
Load850(x0, x1)
Cond_Load850(TRUE, x0, x1)

(8) ITRStoIDPProof (EQUIVALENT transformation)

Added dependency pairs

(9) 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:
Load850(i47, i48) → Cond_Load850(i48 >= 0 && i48 <= 1000 && i48 + 3 > 0, i47, i48)
Cond_Load850(TRUE, i47, i48) → Load850(i47 + i48, i48 + 3)

The integer pair graph contains the following rules and edges:
(0): LOAD850(i47[0], i48[0]) → COND_LOAD850(i48[0] >= 0 && i48[0] <= 1000 && i48[0] + 3 > 0, i47[0], i48[0])
(1): COND_LOAD850(TRUE, i47[1], i48[1]) → LOAD850(i47[1] + i48[1], i48[1] + 3)

(0) -> (1), if ((i48[0] →* i48[1])∧(i47[0] →* i47[1])∧(i48[0] >= 0 && i48[0] <= 1000 && i48[0] + 3 > 0 →* TRUE))


(1) -> (0), if ((i48[1] + 3 →* i48[0])∧(i47[1] + i48[1] →* i47[0]))



The set Q consists of the following terms:
Load850(x0, x1)
Cond_Load850(TRUE, x0, x1)

(10) 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.

(11) 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): LOAD850(i47[0], i48[0]) → COND_LOAD850(i48[0] >= 0 && i48[0] <= 1000 && i48[0] + 3 > 0, i47[0], i48[0])
(1): COND_LOAD850(TRUE, i47[1], i48[1]) → LOAD850(i47[1] + i48[1], i48[1] + 3)

(0) -> (1), if ((i48[0] →* i48[1])∧(i47[0] →* i47[1])∧(i48[0] >= 0 && i48[0] <= 1000 && i48[0] + 3 > 0 →* TRUE))


(1) -> (0), if ((i48[1] + 3 →* i48[0])∧(i47[1] + i48[1] →* i47[0]))



The set Q consists of the following terms:
Load850(x0, x1)
Cond_Load850(TRUE, x0, x1)

(12) 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 LOAD850(i47, i48) → COND_LOAD850(&&(&&(>=(i48, 0), <=(i48, 1000)), >(+(i48, 3), 0)), i47, i48) the following chains were created:
  • We consider the chain LOAD850(i47[0], i48[0]) → COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0]), COND_LOAD850(TRUE, i47[1], i48[1]) → LOAD850(+(i47[1], i48[1]), +(i48[1], 3)) which results in the following constraint:

    (1)    (i48[0]=i48[1]∧i47[0]=i47[1]∧&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0))=TRUE ⇒ LOAD850(i47[0], i48[0])≥NonInfC∧LOAD850(i47[0], i48[0])≥COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])∧(UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥))



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

    (2)    (>(+(i48[0], 3), 0)=TRUE∧>=(i48[0], 0)=TRUE∧<=(i48[0], 1000)=TRUE ⇒ LOAD850(i47[0], i48[0])≥NonInfC∧LOAD850(i47[0], i48[0])≥COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])∧(UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥))



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

    (3)    (i48[0] + [2] ≥ 0∧i48[0] ≥ 0∧[1000] + [-1]i48[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i48[0] ≥ 0∧[(-1)bso_12] ≥ 0)



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

    (4)    (i48[0] + [2] ≥ 0∧i48[0] ≥ 0∧[1000] + [-1]i48[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i48[0] ≥ 0∧[(-1)bso_12] ≥ 0)



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

    (5)    (i48[0] + [2] ≥ 0∧i48[0] ≥ 0∧[1000] + [-1]i48[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥)∧[(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i48[0] ≥ 0∧[(-1)bso_12] ≥ 0)



    We simplified constraint (5) using rules (IDP_UNRESTRICTED_VARS), (IDP_POLY_GCD) which results in the following new constraint:

    (6)    (i48[0] ≥ 0∧[1000] + [-1]i48[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥)∧0 = 0∧[(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i48[0] ≥ 0∧0 = 0∧[(-1)bso_12] ≥ 0)







For Pair COND_LOAD850(TRUE, i47, i48) → LOAD850(+(i47, i48), +(i48, 3)) the following chains were created:
  • We consider the chain COND_LOAD850(TRUE, i47[1], i48[1]) → LOAD850(+(i47[1], i48[1]), +(i48[1], 3)) which results in the following constraint:

    (7)    (COND_LOAD850(TRUE, i47[1], i48[1])≥NonInfC∧COND_LOAD850(TRUE, i47[1], i48[1])≥LOAD850(+(i47[1], i48[1]), +(i48[1], 3))∧(UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥))



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

    (8)    ((UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥)∧[3 + (-1)bso_14] ≥ 0)



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

    (9)    ((UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥)∧[3 + (-1)bso_14] ≥ 0)



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

    (10)    ((UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥)∧[3 + (-1)bso_14] ≥ 0)



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

    (11)    ((UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥)∧0 = 0∧0 = 0∧[3 + (-1)bso_14] ≥ 0)







To summarize, we get the following constraints P≥ for the following pairs.
  • LOAD850(i47, i48) → COND_LOAD850(&&(&&(>=(i48, 0), <=(i48, 1000)), >(+(i48, 3), 0)), i47, i48)
    • (i48[0] ≥ 0∧[1000] + [-1]i48[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])), ≥)∧0 = 0∧[(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]i48[0] ≥ 0∧0 = 0∧[(-1)bso_12] ≥ 0)

  • COND_LOAD850(TRUE, i47, i48) → LOAD850(+(i47, i48), +(i48, 3))
    • ((UIncreasing(LOAD850(+(i47[1], i48[1]), +(i48[1], 3))), ≥)∧0 = 0∧0 = 0∧[3 + (-1)bso_14] ≥ 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(LOAD850(x1, x2)) = [-1] + [-1]x2   
POL(COND_LOAD850(x1, x2, x3)) = [-1] + [-1]x3   
POL(&&(x1, x2)) = [-1]   
POL(>=(x1, x2)) = [-1]   
POL(0) = 0   
POL(<=(x1, x2)) = [-1]   
POL(1000) = [1000]   
POL(>(x1, x2)) = [-1]   
POL(+(x1, x2)) = x1 + x2   
POL(3) = [3]   

The following pairs are in P>:

COND_LOAD850(TRUE, i47[1], i48[1]) → LOAD850(+(i47[1], i48[1]), +(i48[1], 3))

The following pairs are in Pbound:

LOAD850(i47[0], i48[0]) → COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])

The following pairs are in P≥:

LOAD850(i47[0], i48[0]) → COND_LOAD850(&&(&&(>=(i48[0], 0), <=(i48[0], 1000)), >(+(i48[0], 3), 0)), i47[0], i48[0])

There are no usable rules.

(13) Complex Obligation (AND)

(14) 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): LOAD850(i47[0], i48[0]) → COND_LOAD850(i48[0] >= 0 && i48[0] <= 1000 && i48[0] + 3 > 0, i47[0], i48[0])


The set Q consists of the following terms:
Load850(x0, x1)
Cond_Load850(TRUE, x0, x1)

(15) IDependencyGraphProof (EQUIVALENT transformation)

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

(16) TRUE

(17) 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_LOAD850(TRUE, i47[1], i48[1]) → LOAD850(i47[1] + i48[1], i48[1] + 3)


The set Q consists of the following terms:
Load850(x0, x1)
Cond_Load850(TRUE, x0, x1)

(18) IDependencyGraphProof (EQUIVALENT transformation)

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

(19) TRUE

(20) 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:
Load553(1000, 1000, 1000, i20, i21) → Cond_Load553(i21 >= 0 && i21 <= 1000 && i21 + 2 > 0, 1000, 1000, 1000, i20, i21)
Cond_Load553(TRUE, 1000, 1000, 1000, i20, i21) → Load553(1000, 1000, 1000, i20 + i21, i21 + 2)
The set Q consists of the following terms:
Load553(1000, 1000, 1000, x0, x1)
Cond_Load553(TRUE, 1000, 1000, 1000, x0, x1)

(21) GroundTermsRemoverProof (EQUIVALENT transformation)

Some arguments are removed because they always contain the same ground term.
We removed the following ground terms:
  • 1000

We removed arguments according to the following replacements:

Load553(x1, x2, x3, x4, x5) → Load553(x4, x5)
Cond_Load553(x1, x2, x3, x4, x5, x6) → Cond_Load553(x1, x5, x6)

(22) 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:
Load553(i20, i21) → Cond_Load553(i21 >= 0 && i21 <= 1000 && i21 + 2 > 0, i20, i21)
Cond_Load553(TRUE, i20, i21) → Load553(i20 + i21, i21 + 2)
The set Q consists of the following terms:
Load553(x0, x1)
Cond_Load553(TRUE, x0, x1)

(23) ITRStoIDPProof (EQUIVALENT transformation)

Added dependency pairs

(24) 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:
Load553(i20, i21) → Cond_Load553(i21 >= 0 && i21 <= 1000 && i21 + 2 > 0, i20, i21)
Cond_Load553(TRUE, i20, i21) → Load553(i20 + i21, i21 + 2)

The integer pair graph contains the following rules and edges:
(0): LOAD553(i20[0], i21[0]) → COND_LOAD553(i21[0] >= 0 && i21[0] <= 1000 && i21[0] + 2 > 0, i20[0], i21[0])
(1): COND_LOAD553(TRUE, i20[1], i21[1]) → LOAD553(i20[1] + i21[1], i21[1] + 2)

(0) -> (1), if ((i21[0] →* i21[1])∧(i20[0] →* i20[1])∧(i21[0] >= 0 && i21[0] <= 1000 && i21[0] + 2 > 0 →* TRUE))


(1) -> (0), if ((i20[1] + i21[1] →* i20[0])∧(i21[1] + 2 →* i21[0]))



The set Q consists of the following terms:
Load553(x0, x1)
Cond_Load553(TRUE, x0, x1)

(25) 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.

(26) 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): LOAD553(i20[0], i21[0]) → COND_LOAD553(i21[0] >= 0 && i21[0] <= 1000 && i21[0] + 2 > 0, i20[0], i21[0])
(1): COND_LOAD553(TRUE, i20[1], i21[1]) → LOAD553(i20[1] + i21[1], i21[1] + 2)

(0) -> (1), if ((i21[0] →* i21[1])∧(i20[0] →* i20[1])∧(i21[0] >= 0 && i21[0] <= 1000 && i21[0] + 2 > 0 →* TRUE))


(1) -> (0), if ((i20[1] + i21[1] →* i20[0])∧(i21[1] + 2 →* i21[0]))



The set Q consists of the following terms:
Load553(x0, x1)
Cond_Load553(TRUE, x0, x1)

(27) 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 LOAD553(i20, i21) → COND_LOAD553(&&(&&(>=(i21, 0), <=(i21, 1000)), >(+(i21, 2), 0)), i20, i21) the following chains were created:
  • We consider the chain LOAD553(i20[0], i21[0]) → COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0]), COND_LOAD553(TRUE, i20[1], i21[1]) → LOAD553(+(i20[1], i21[1]), +(i21[1], 2)) which results in the following constraint:

    (1)    (i21[0]=i21[1]∧i20[0]=i20[1]∧&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0))=TRUE ⇒ LOAD553(i20[0], i21[0])≥NonInfC∧LOAD553(i20[0], i21[0])≥COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])∧(UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥))



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

    (2)    (>(+(i21[0], 2), 0)=TRUE∧>=(i21[0], 0)=TRUE∧<=(i21[0], 1000)=TRUE ⇒ LOAD553(i20[0], i21[0])≥NonInfC∧LOAD553(i20[0], i21[0])≥COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])∧(UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥))



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

    (3)    (i21[0] + [1] ≥ 0∧i21[0] ≥ 0∧[1000] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i21[0] ≥ 0∧[1 + (-1)bso_12] ≥ 0)



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

    (4)    (i21[0] + [1] ≥ 0∧i21[0] ≥ 0∧[1000] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i21[0] ≥ 0∧[1 + (-1)bso_12] ≥ 0)



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

    (5)    (i21[0] + [1] ≥ 0∧i21[0] ≥ 0∧[1000] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i21[0] ≥ 0∧[1 + (-1)bso_12] ≥ 0)



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

    (6)    (i21[0] + [1] ≥ 0∧i21[0] ≥ 0∧[1000] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥)∧0 = 0∧[(-1)Bound*bni_11] + [(-1)bni_11]i21[0] ≥ 0∧0 = 0∧[1 + (-1)bso_12] ≥ 0)







For Pair COND_LOAD553(TRUE, i20, i21) → LOAD553(+(i20, i21), +(i21, 2)) the following chains were created:
  • We consider the chain COND_LOAD553(TRUE, i20[1], i21[1]) → LOAD553(+(i20[1], i21[1]), +(i21[1], 2)) which results in the following constraint:

    (7)    (COND_LOAD553(TRUE, i20[1], i21[1])≥NonInfC∧COND_LOAD553(TRUE, i20[1], i21[1])≥LOAD553(+(i20[1], i21[1]), +(i21[1], 2))∧(UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥))



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

    (8)    ((UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥)∧[1 + (-1)bso_14] ≥ 0)



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

    (9)    ((UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥)∧[1 + (-1)bso_14] ≥ 0)



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

    (10)    ((UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥)∧[1 + (-1)bso_14] ≥ 0)



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

    (11)    ((UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_14] ≥ 0)







To summarize, we get the following constraints P≥ for the following pairs.
  • LOAD553(i20, i21) → COND_LOAD553(&&(&&(>=(i21, 0), <=(i21, 1000)), >(+(i21, 2), 0)), i20, i21)
    • (i21[0] + [1] ≥ 0∧i21[0] ≥ 0∧[1000] + [-1]i21[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])), ≥)∧0 = 0∧[(-1)Bound*bni_11] + [(-1)bni_11]i21[0] ≥ 0∧0 = 0∧[1 + (-1)bso_12] ≥ 0)

  • COND_LOAD553(TRUE, i20, i21) → LOAD553(+(i20, i21), +(i21, 2))
    • ((UIncreasing(LOAD553(+(i20[1], i21[1]), +(i21[1], 2))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_14] ≥ 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(LOAD553(x1, x2)) = [-1]x2   
POL(COND_LOAD553(x1, x2, x3)) = [-1] + [-1]x3   
POL(&&(x1, x2)) = [-1]   
POL(>=(x1, x2)) = [-1]   
POL(0) = 0   
POL(<=(x1, x2)) = [-1]   
POL(1000) = [1000]   
POL(>(x1, x2)) = [-1]   
POL(+(x1, x2)) = x1 + x2   
POL(2) = [2]   

The following pairs are in P>:

LOAD553(i20[0], i21[0]) → COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])
COND_LOAD553(TRUE, i20[1], i21[1]) → LOAD553(+(i20[1], i21[1]), +(i21[1], 2))

The following pairs are in Pbound:

LOAD553(i20[0], i21[0]) → COND_LOAD553(&&(&&(>=(i21[0], 0), <=(i21[0], 1000)), >(+(i21[0], 2), 0)), i20[0], i21[0])

The following pairs are in P≥:
none

There are no usable rules.

(28) Complex Obligation (AND)

(29) 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:
Load553(x0, x1)
Cond_Load553(TRUE, x0, x1)

(30) IDependencyGraphProof (EQUIVALENT transformation)

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

(31) TRUE

(32) 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_LOAD553(TRUE, i20[1], i21[1]) → LOAD553(i20[1] + i21[1], i21[1] + 2)


The set Q consists of the following terms:
Load553(x0, x1)
Cond_Load553(TRUE, x0, x1)

(33) IDependencyGraphProof (EQUIVALENT transformation)

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

(34) TRUE

(35) 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:
Load257(1000, 1000, 1000, i5, i6) → Cond_Load257(i6 >= 0 && i6 <= 1000 && i6 + 1 > 0, 1000, 1000, 1000, i5, i6)
Cond_Load257(TRUE, 1000, 1000, 1000, i5, i6) → Load257(1000, 1000, 1000, i5 + i6, i6 + 1)
The set Q consists of the following terms:
Load257(1000, 1000, 1000, x0, x1)
Cond_Load257(TRUE, 1000, 1000, 1000, x0, x1)

(36) GroundTermsRemoverProof (EQUIVALENT transformation)

Some arguments are removed because they always contain the same ground term.
We removed the following ground terms:
  • 1000

We removed arguments according to the following replacements:

Load257(x1, x2, x3, x4, x5) → Load257(x4, x5)
Cond_Load257(x1, x2, x3, x4, x5, x6) → Cond_Load257(x1, x5, x6)

(37) 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:
Load257(i5, i6) → Cond_Load257(i6 >= 0 && i6 <= 1000 && i6 + 1 > 0, i5, i6)
Cond_Load257(TRUE, i5, i6) → Load257(i5 + i6, i6 + 1)
The set Q consists of the following terms:
Load257(x0, x1)
Cond_Load257(TRUE, x0, x1)

(38) ITRStoIDPProof (EQUIVALENT transformation)

Added dependency pairs

(39) 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:
Load257(i5, i6) → Cond_Load257(i6 >= 0 && i6 <= 1000 && i6 + 1 > 0, i5, i6)
Cond_Load257(TRUE, i5, i6) → Load257(i5 + i6, i6 + 1)

The integer pair graph contains the following rules and edges:
(0): LOAD257(i5[0], i6[0]) → COND_LOAD257(i6[0] >= 0 && i6[0] <= 1000 && i6[0] + 1 > 0, i5[0], i6[0])
(1): COND_LOAD257(TRUE, i5[1], i6[1]) → LOAD257(i5[1] + i6[1], i6[1] + 1)

(0) -> (1), if ((i6[0] >= 0 && i6[0] <= 1000 && i6[0] + 1 > 0 →* TRUE)∧(i6[0] →* i6[1])∧(i5[0] →* i5[1]))


(1) -> (0), if ((i5[1] + i6[1] →* i5[0])∧(i6[1] + 1 →* i6[0]))



The set Q consists of the following terms:
Load257(x0, x1)
Cond_Load257(TRUE, x0, x1)

(40) 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.

(41) 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): LOAD257(i5[0], i6[0]) → COND_LOAD257(i6[0] >= 0 && i6[0] <= 1000 && i6[0] + 1 > 0, i5[0], i6[0])
(1): COND_LOAD257(TRUE, i5[1], i6[1]) → LOAD257(i5[1] + i6[1], i6[1] + 1)

(0) -> (1), if ((i6[0] >= 0 && i6[0] <= 1000 && i6[0] + 1 > 0 →* TRUE)∧(i6[0] →* i6[1])∧(i5[0] →* i5[1]))


(1) -> (0), if ((i5[1] + i6[1] →* i5[0])∧(i6[1] + 1 →* i6[0]))



The set Q consists of the following terms:
Load257(x0, x1)
Cond_Load257(TRUE, x0, x1)

(42) 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 LOAD257(i5, i6) → COND_LOAD257(&&(&&(>=(i6, 0), <=(i6, 1000)), >(+(i6, 1), 0)), i5, i6) the following chains were created:
  • We consider the chain LOAD257(i5[0], i6[0]) → COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0]), COND_LOAD257(TRUE, i5[1], i6[1]) → LOAD257(+(i5[1], i6[1]), +(i6[1], 1)) which results in the following constraint:

    (1)    (&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0))=TRUE∧i6[0]=i6[1]∧i5[0]=i5[1] ⇒ LOAD257(i5[0], i6[0])≥NonInfC∧LOAD257(i5[0], i6[0])≥COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])∧(UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥))



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

    (2)    (>(+(i6[0], 1), 0)=TRUE∧>=(i6[0], 0)=TRUE∧<=(i6[0], 1000)=TRUE ⇒ LOAD257(i5[0], i6[0])≥NonInfC∧LOAD257(i5[0], i6[0])≥COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])∧(UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥))



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

    (3)    (i6[0] ≥ 0∧i6[0] ≥ 0∧[1000] + [-1]i6[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i6[0] ≥ 0∧[(-1)bso_12] ≥ 0)



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

    (4)    (i6[0] ≥ 0∧i6[0] ≥ 0∧[1000] + [-1]i6[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i6[0] ≥ 0∧[(-1)bso_12] ≥ 0)



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

    (5)    (i6[0] ≥ 0∧i6[0] ≥ 0∧[1000] + [-1]i6[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥)∧[(-1)Bound*bni_11] + [(-1)bni_11]i6[0] ≥ 0∧[(-1)bso_12] ≥ 0)



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

    (6)    (i6[0] ≥ 0∧i6[0] ≥ 0∧[1000] + [-1]i6[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥)∧0 = 0∧[(-1)Bound*bni_11] + [(-1)bni_11]i6[0] ≥ 0∧0 = 0∧[(-1)bso_12] ≥ 0)







For Pair COND_LOAD257(TRUE, i5, i6) → LOAD257(+(i5, i6), +(i6, 1)) the following chains were created:
  • We consider the chain COND_LOAD257(TRUE, i5[1], i6[1]) → LOAD257(+(i5[1], i6[1]), +(i6[1], 1)) which results in the following constraint:

    (7)    (COND_LOAD257(TRUE, i5[1], i6[1])≥NonInfC∧COND_LOAD257(TRUE, i5[1], i6[1])≥LOAD257(+(i5[1], i6[1]), +(i6[1], 1))∧(UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥))



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

    (8)    ((UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥)∧[1 + (-1)bso_14] ≥ 0)



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

    (9)    ((UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥)∧[1 + (-1)bso_14] ≥ 0)



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

    (10)    ((UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥)∧[1 + (-1)bso_14] ≥ 0)



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

    (11)    ((UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_14] ≥ 0)







To summarize, we get the following constraints P≥ for the following pairs.
  • LOAD257(i5, i6) → COND_LOAD257(&&(&&(>=(i6, 0), <=(i6, 1000)), >(+(i6, 1), 0)), i5, i6)
    • (i6[0] ≥ 0∧i6[0] ≥ 0∧[1000] + [-1]i6[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])), ≥)∧0 = 0∧[(-1)Bound*bni_11] + [(-1)bni_11]i6[0] ≥ 0∧0 = 0∧[(-1)bso_12] ≥ 0)

  • COND_LOAD257(TRUE, i5, i6) → LOAD257(+(i5, i6), +(i6, 1))
    • ((UIncreasing(LOAD257(+(i5[1], i6[1]), +(i6[1], 1))), ≥)∧0 = 0∧0 = 0∧[1 + (-1)bso_14] ≥ 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(LOAD257(x1, x2)) = [-1]x2   
POL(COND_LOAD257(x1, x2, x3)) = [-1]x3   
POL(&&(x1, x2)) = [-1]   
POL(>=(x1, x2)) = [-1]   
POL(0) = 0   
POL(<=(x1, x2)) = [-1]   
POL(1000) = [1000]   
POL(>(x1, x2)) = [-1]   
POL(+(x1, x2)) = x1 + x2   
POL(1) = [1]   

The following pairs are in P>:

COND_LOAD257(TRUE, i5[1], i6[1]) → LOAD257(+(i5[1], i6[1]), +(i6[1], 1))

The following pairs are in Pbound:

LOAD257(i5[0], i6[0]) → COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])

The following pairs are in P≥:

LOAD257(i5[0], i6[0]) → COND_LOAD257(&&(&&(>=(i6[0], 0), <=(i6[0], 1000)), >(+(i6[0], 1), 0)), i5[0], i6[0])

There are no usable rules.

(43) Complex Obligation (AND)

(44) 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): LOAD257(i5[0], i6[0]) → COND_LOAD257(i6[0] >= 0 && i6[0] <= 1000 && i6[0] + 1 > 0, i5[0], i6[0])


The set Q consists of the following terms:
Load257(x0, x1)
Cond_Load257(TRUE, x0, x1)

(45) IDependencyGraphProof (EQUIVALENT transformation)

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

(46) TRUE

(47) 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_LOAD257(TRUE, i5[1], i6[1]) → LOAD257(i5[1] + i6[1], i6[1] + 1)


The set Q consists of the following terms:
Load257(x0, x1)
Cond_Load257(TRUE, x0, x1)

(48) IDependencyGraphProof (EQUIVALENT transformation)

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

(49) TRUE