Termination of the given JBC could be proven:

0 JBC
↳1 JBC2FIG (⇐)
↳2 FIGraph
↳3 FIGtoITRSProof (⇐)
↳4 ITRS
↳5 ITRStoIDPProof (⇔)
↳6 IDP
↳7 UsableRulesProof (⇔)
↳8 IDP
↳9 IDPNonInfProof (⇐)
↳10 AND
    ↳11 IDP
        ↳12 IDependencyGraphProof (⇔)
        ↳13 TRUE
    ↳14 IDP
        ↳15 IDependencyGraphProof (⇔)
        ↳16 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: PastaC7

(1) JBC2FIG (SOUND transformation)

Constructed FIGraph.

(2) Obligation:

FIGraph based on JBC Program:
Graph of 237 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:
Load1135(i101, i102, i103) → Cond_Load1135(i102 >= 0 && i102 <= i103 && i101 >= 0 && i101 <= 100 && i102 + 1 > 0, i101, i102, i103)
Cond_Load1135(TRUE, i101, i102, i103) → Load1135(i102, i102 + 1, i103 + -1)
The set Q consists of the following terms:
Load1135(x0, x1, x2)
Cond_Load1135(TRUE, x0, x1, x2)

(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:

Boolean, Integer


The ITRS R consists of the following rules:
Load1135(i101, i102, i103) → Cond_Load1135(i102 >= 0 && i102 <= i103 && i101 >= 0 && i101 <= 100 && i102 + 1 > 0, i101, i102, i103)
Cond_Load1135(TRUE, i101, i102, i103) → Load1135(i102, i102 + 1, i103 + -1)

The integer pair graph contains the following rules and edges:
(0): LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(i102[0] >= 0 && i102[0] <= i103[0] && i101[0] >= 0 && i101[0] <= 100 && i102[0] + 1 > 0, i101[0], i102[0], i103[0])
(1): COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], i102[1] + 1, i103[1] + -1)

(0) -> (1), if ((i101[0] →* i101[1])∧(i102[0] →* i102[1])∧(i102[0] >= 0 && i102[0] <= i103[0] && i101[0] >= 0 && i101[0] <= 100 && i102[0] + 1 > 0 →* TRUE)∧(i103[0] →* i103[1]))


(1) -> (0), if ((i102[1] →* i101[0])∧(i103[1] + -1 →* i103[0])∧(i102[1] + 1 →* i102[0]))



The set Q consists of the following terms:
Load1135(x0, x1, x2)
Cond_Load1135(TRUE, x0, x1, x2)

(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): LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(i102[0] >= 0 && i102[0] <= i103[0] && i101[0] >= 0 && i101[0] <= 100 && i102[0] + 1 > 0, i101[0], i102[0], i103[0])
(1): COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], i102[1] + 1, i103[1] + -1)

(0) -> (1), if ((i101[0] →* i101[1])∧(i102[0] →* i102[1])∧(i102[0] >= 0 && i102[0] <= i103[0] && i101[0] >= 0 && i101[0] <= 100 && i102[0] + 1 > 0 →* TRUE)∧(i103[0] →* i103[1]))


(1) -> (0), if ((i102[1] →* i101[0])∧(i103[1] + -1 →* i103[0])∧(i102[1] + 1 →* i102[0]))



The set Q consists of the following terms:
Load1135(x0, x1, x2)
Cond_Load1135(TRUE, x0, x1, x2)

(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 LOAD1135(i101, i102, i103) → COND_LOAD1135(&&(&&(&&(&&(>=(i102, 0), <=(i102, i103)), >=(i101, 0)), <=(i101, 100)), >(+(i102, 1), 0)), i101, i102, i103) the following chains were created:
  • We consider the chain LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0]), COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1)) which results in the following constraint:

    (1)    (i101[0]=i101[1]∧i102[0]=i102[1]∧&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0))=TRUE∧i103[0]=i103[1] ⇒ LOAD1135(i101[0], i102[0], i103[0])≥NonInfC∧LOAD1135(i101[0], i102[0], i103[0])≥COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])∧(UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥))



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

    (2)    (>(+(i102[0], 1), 0)=TRUE∧<=(i101[0], 100)=TRUE∧>=(i101[0], 0)=TRUE∧>=(i102[0], 0)=TRUE∧<=(i102[0], i103[0])=TRUE ⇒ LOAD1135(i101[0], i102[0], i103[0])≥NonInfC∧LOAD1135(i101[0], i102[0], i103[0])≥COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])∧(UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥))



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

    (3)    (i102[0] ≥ 0∧[100] + [-1]i101[0] ≥ 0∧i101[0] ≥ 0∧i102[0] ≥ 0∧i103[0] + [-1]i102[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥)∧[bni_12 + (-1)Bound*bni_12] + [bni_12]i103[0] + [(-1)bni_12]i102[0] ≥ 0∧[(-1)bso_13] ≥ 0)



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

    (4)    (i102[0] ≥ 0∧[100] + [-1]i101[0] ≥ 0∧i101[0] ≥ 0∧i102[0] ≥ 0∧i103[0] + [-1]i102[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥)∧[bni_12 + (-1)Bound*bni_12] + [bni_12]i103[0] + [(-1)bni_12]i102[0] ≥ 0∧[(-1)bso_13] ≥ 0)



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

    (5)    (i102[0] ≥ 0∧[100] + [-1]i101[0] ≥ 0∧i101[0] ≥ 0∧i102[0] ≥ 0∧i103[0] + [-1]i102[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥)∧[bni_12 + (-1)Bound*bni_12] + [bni_12]i103[0] + [(-1)bni_12]i102[0] ≥ 0∧[(-1)bso_13] ≥ 0)



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

    (6)    (i102[0] ≥ 0∧[100] + [-1]i101[0] ≥ 0∧i101[0] ≥ 0∧i102[0] ≥ 0∧i103[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥)∧[bni_12 + (-1)Bound*bni_12] + [bni_12]i103[0] ≥ 0∧[(-1)bso_13] ≥ 0)







For Pair COND_LOAD1135(TRUE, i101, i102, i103) → LOAD1135(i102, +(i102, 1), +(i103, -1)) the following chains were created:
  • We consider the chain COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1)) which results in the following constraint:

    (7)    (COND_LOAD1135(TRUE, i101[1], i102[1], i103[1])≥NonInfC∧COND_LOAD1135(TRUE, i101[1], i102[1], i103[1])≥LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))∧(UIncreasing(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥))



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

    (8)    ((UIncreasing(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥)∧[2 + (-1)bso_15] ≥ 0)



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

    (9)    ((UIncreasing(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥)∧[2 + (-1)bso_15] ≥ 0)



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

    (10)    ((UIncreasing(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥)∧[2 + (-1)bso_15] ≥ 0)



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

    (11)    ((UIncreasing(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥)∧0 = 0∧0 = 0∧0 = 0∧[2 + (-1)bso_15] ≥ 0)







To summarize, we get the following constraints P≥ for the following pairs.
  • LOAD1135(i101, i102, i103) → COND_LOAD1135(&&(&&(&&(&&(>=(i102, 0), <=(i102, i103)), >=(i101, 0)), <=(i101, 100)), >(+(i102, 1), 0)), i101, i102, i103)
    • (i102[0] ≥ 0∧[100] + [-1]i101[0] ≥ 0∧i101[0] ≥ 0∧i102[0] ≥ 0∧i103[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])), ≥)∧[bni_12 + (-1)Bound*bni_12] + [bni_12]i103[0] ≥ 0∧[(-1)bso_13] ≥ 0)

  • COND_LOAD1135(TRUE, i101, i102, i103) → LOAD1135(i102, +(i102, 1), +(i103, -1))
    • ((UIncreasing(LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))), ≥)∧0 = 0∧0 = 0∧0 = 0∧[2 + (-1)bso_15] ≥ 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(LOAD1135(x1, x2, x3)) = [1] + x3 + [-1]x2   
POL(COND_LOAD1135(x1, x2, x3, x4)) = [1] + x4 + [-1]x3   
POL(&&(x1, x2)) = [-1]   
POL(>=(x1, x2)) = [-1]   
POL(0) = 0   
POL(<=(x1, x2)) = [-1]   
POL(100) = [100]   
POL(>(x1, x2)) = [-1]   
POL(+(x1, x2)) = x1 + x2   
POL(1) = [1]   
POL(-1) = [-1]   

The following pairs are in P>:

COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], +(i102[1], 1), +(i103[1], -1))

The following pairs are in Pbound:

LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])

The following pairs are in P≥:

LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(&&(&&(&&(&&(>=(i102[0], 0), <=(i102[0], i103[0])), >=(i101[0], 0)), <=(i101[0], 100)), >(+(i102[0], 1), 0)), i101[0], i102[0], i103[0])

There are no usable rules.

(10) Complex Obligation (AND)

(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): LOAD1135(i101[0], i102[0], i103[0]) → COND_LOAD1135(i102[0] >= 0 && i102[0] <= i103[0] && i101[0] >= 0 && i101[0] <= 100 && i102[0] + 1 > 0, i101[0], i102[0], i103[0])


The set Q consists of the following terms:
Load1135(x0, x1, x2)
Cond_Load1135(TRUE, x0, x1, x2)

(12) IDependencyGraphProof (EQUIVALENT transformation)

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

(13) TRUE

(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:

Integer


R is empty.

The integer pair graph contains the following rules and edges:
(1): COND_LOAD1135(TRUE, i101[1], i102[1], i103[1]) → LOAD1135(i102[1], i102[1] + 1, i103[1] + -1)


The set Q consists of the following terms:
Load1135(x0, x1, x2)
Cond_Load1135(TRUE, x0, x1, x2)

(15) IDependencyGraphProof (EQUIVALENT transformation)

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

(16) TRUE