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

public class A {

	int incr(int i) {
		return i=i+1;
	}
}


package example_1;

public class B extends A {
	int incr(int i) {
		return i = i+2;
	}
}



package example_1;

public class C extends B {
	int incr(int i) {
		return i=i+3;
	}
}


package example_1;

public class Test {

	public int add(int n,A o){
		int res=0;
		int i=0;
		while (i<=n){
			res=res+i;
			i=o.incr(i);
		}    
		return res;
	}

	public static void main(String[] args) {
		int test = 1000;
		Test testClass = new Test();
		A a = new A();
		int result1 = testClass.add(test,a);
		a = new B();
		int result2 = testClass.add(test,a);
		a = new C(); 
		int result3 = testClass.add(test,a);     
		// System.out.println("Result: "+result1 + result2 + result3);
	}
}


(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:
Load1041(1000, 1000, i46, i47) → Cond_Load1041(i47 >= 0 && i47 <= 1000 && i47 + 3 > 0, 1000, 1000, i46, i47)
Cond_Load1041(TRUE, 1000, 1000, i46, i47) → Load1041(1000, 1000, i46 + i47, i47 + 3)
The set Q consists of the following terms:
Load1041(1000, 1000, x0, x1)
Cond_Load1041(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:

Load1041(x1, x2, x3, x4) → Load1041(x3, x4)
Cond_Load1041(x1, x2, x3, x4, x5) → Cond_Load1041(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:
Load1041(i46, i47) → Cond_Load1041(i47 >= 0 && i47 <= 1000 && i47 + 3 > 0, i46, i47)
Cond_Load1041(TRUE, i46, i47) → Load1041(i46 + i47, i47 + 3)
The set Q consists of the following terms:
Load1041(x0, x1)
Cond_Load1041(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:
Load1041(i46, i47) → Cond_Load1041(i47 >= 0 && i47 <= 1000 && i47 + 3 > 0, i46, i47)
Cond_Load1041(TRUE, i46, i47) → Load1041(i46 + i47, i47 + 3)

The integer pair graph contains the following rules and edges:
(0): LOAD1041(i46[0], i47[0]) → COND_LOAD1041(i47[0] >= 0 && i47[0] <= 1000 && i47[0] + 3 > 0, i46[0], i47[0])
(1): COND_LOAD1041(TRUE, i46[1], i47[1]) → LOAD1041(i46[1] + i47[1], i47[1] + 3)

(0) -> (1), if ((i47[0] →* i47[1])∧(i46[0] →* i46[1])∧(i47[0] >= 0 && i47[0] <= 1000 && i47[0] + 3 > 0 →* TRUE))


(1) -> (0), if ((i47[1] + 3 →* i47[0])∧(i46[1] + i47[1] →* i46[0]))



The set Q consists of the following terms:
Load1041(x0, x1)
Cond_Load1041(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): LOAD1041(i46[0], i47[0]) → COND_LOAD1041(i47[0] >= 0 && i47[0] <= 1000 && i47[0] + 3 > 0, i46[0], i47[0])
(1): COND_LOAD1041(TRUE, i46[1], i47[1]) → LOAD1041(i46[1] + i47[1], i47[1] + 3)

(0) -> (1), if ((i47[0] →* i47[1])∧(i46[0] →* i46[1])∧(i47[0] >= 0 && i47[0] <= 1000 && i47[0] + 3 > 0 →* TRUE))


(1) -> (0), if ((i47[1] + 3 →* i47[0])∧(i46[1] + i47[1] →* i46[0]))



The set Q consists of the following terms:
Load1041(x0, x1)
Cond_Load1041(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 LOAD1041(i46, i47) → COND_LOAD1041(&&(&&(>=(i47, 0), <=(i47, 1000)), >(+(i47, 3), 0)), i46, i47) the following chains were created:
  • We consider the chain LOAD1041(i46[0], i47[0]) → COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0]), COND_LOAD1041(TRUE, i46[1], i47[1]) → LOAD1041(+(i46[1], i47[1]), +(i47[1], 3)) which results in the following constraint:

    (1)    (i47[0]=i47[1]∧i46[0]=i46[1]∧&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0))=TRUE ⇒ LOAD1041(i46[0], i47[0])≥NonInfC∧LOAD1041(i46[0], i47[0])≥COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])∧(UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥))



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

    (2)    (>(+(i47[0], 3), 0)=TRUE∧>=(i47[0], 0)=TRUE∧<=(i47[0], 1000)=TRUE ⇒ LOAD1041(i46[0], i47[0])≥NonInfC∧LOAD1041(i46[0], i47[0])≥COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])∧(UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥))



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

    (3)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i47[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (4)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i47[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (5)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i47[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (6)    (i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥)∧0 = 0∧[(-1)Bound*bni_14] + [(-1)bni_14]i47[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)







For Pair COND_LOAD1041(TRUE, i46, i47) → LOAD1041(+(i46, i47), +(i47, 3)) the following chains were created:
  • We consider the chain LOAD1041(i46[0], i47[0]) → COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0]), COND_LOAD1041(TRUE, i46[1], i47[1]) → LOAD1041(+(i46[1], i47[1]), +(i47[1], 3)), LOAD1041(i46[0], i47[0]) → COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0]) which results in the following constraint:

    (7)    (i47[0]=i47[1]∧i46[0]=i46[1]∧&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0))=TRUE∧+(i47[1], 3)=i47[0]1∧+(i46[1], i47[1])=i46[0]1 ⇒ COND_LOAD1041(TRUE, i46[1], i47[1])≥NonInfC∧COND_LOAD1041(TRUE, i46[1], i47[1])≥LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))∧(UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥))



    We simplified constraint (7) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:

    (8)    (>(+(i47[0], 3), 0)=TRUE∧>=(i47[0], 0)=TRUE∧<=(i47[0], 1000)=TRUE ⇒ COND_LOAD1041(TRUE, i46[0], i47[0])≥NonInfC∧COND_LOAD1041(TRUE, i46[0], i47[0])≥LOAD1041(+(i46[0], i47[0]), +(i47[0], 3))∧(UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥))



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

    (9)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i47[0] ≥ 0∧[3 + (-1)bso_17] ≥ 0)



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

    (10)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i47[0] ≥ 0∧[3 + (-1)bso_17] ≥ 0)



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

    (11)    (i47[0] + [2] ≥ 0∧i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0 ⇒ (UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i47[0] ≥ 0∧[3 + (-1)bso_17] ≥ 0)



    We simplified constraint (11) using rules (IDP_UNRESTRICTED_VARS), (IDP_POLY_GCD) which results in the following new constraint:

    (12)    (i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥)∧0 = 0∧[(-1)Bound*bni_16] + [(-1)bni_16]i47[0] ≥ 0∧0 = 0∧[3 + (-1)bso_17] ≥ 0)







To summarize, we get the following constraints P≥ for the following pairs.
  • LOAD1041(i46, i47) → COND_LOAD1041(&&(&&(>=(i47, 0), <=(i47, 1000)), >(+(i47, 3), 0)), i46, i47)
    • (i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])), ≥)∧0 = 0∧[(-1)Bound*bni_14] + [(-1)bni_14]i47[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)

  • COND_LOAD1041(TRUE, i46, i47) → LOAD1041(+(i46, i47), +(i47, 3))
    • (i47[0] ≥ 0∧[1000] + [-1]i47[0] ≥ 0∧[1] ≥ 0 ⇒ (UIncreasing(LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))), ≥)∧0 = 0∧[(-1)Bound*bni_16] + [(-1)bni_16]i47[0] ≥ 0∧0 = 0∧[3 + (-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) = [1]   
POL(LOAD1041(x1, x2)) = [-1]x2   
POL(COND_LOAD1041(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(3) = [3]   

The following pairs are in P>:

COND_LOAD1041(TRUE, i46[1], i47[1]) → LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))

The following pairs are in Pbound:

LOAD1041(i46[0], i47[0]) → COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])
COND_LOAD1041(TRUE, i46[1], i47[1]) → LOAD1041(+(i46[1], i47[1]), +(i47[1], 3))

The following pairs are in P≥:

LOAD1041(i46[0], i47[0]) → COND_LOAD1041(&&(&&(>=(i47[0], 0), <=(i47[0], 1000)), >(+(i47[0], 3), 0)), i46[0], i47[0])

At least the following rules have been oriented under context sensitive arithmetic replacement:

FALSE1 → &&(FALSE, FALSE)1

(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): LOAD1041(i46[0], i47[0]) → COND_LOAD1041(i47[0] >= 0 && i47[0] <= 1000 && i47[0] + 3 > 0, i46[0], i47[0])


The set Q consists of the following terms:
Load1041(x0, x1)
Cond_Load1041(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:
none


R is empty.

The integer pair graph is empty.

The set Q consists of the following terms:
Load1041(x0, x1)
Cond_Load1041(TRUE, x0, x1)

(18) IDependencyGraphProof (EQUIVALENT transformation)

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

(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:
Load731(1000, 1000, 1000, i19, i20) → Cond_Load731(i20 >= 0 && i20 <= 1000 && i20 + 2 > 0, 1000, 1000, 1000, i19, i20)
Cond_Load731(TRUE, 1000, 1000, 1000, i19, i20) → Load731(1000, 1000, 1000, i19 + i20, i20 + 2)
The set Q consists of the following terms:
Load731(1000, 1000, 1000, x0, x1)
Cond_Load731(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:

Load731(x1, x2, x3, x4, x5) → Load731(x4, x5)
Cond_Load731(x1, x2, x3, x4, x5, x6) → Cond_Load731(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:
Load731(i19, i20) → Cond_Load731(i20 >= 0 && i20 <= 1000 && i20 + 2 > 0, i19, i20)
Cond_Load731(TRUE, i19, i20) → Load731(i19 + i20, i20 + 2)
The set Q consists of the following terms:
Load731(x0, x1)
Cond_Load731(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:
Load731(i19, i20) → Cond_Load731(i20 >= 0 && i20 <= 1000 && i20 + 2 > 0, i19, i20)
Cond_Load731(TRUE, i19, i20) → Load731(i19 + i20, i20 + 2)

The integer pair graph contains the following rules and edges:
(0): LOAD731(i19[0], i20[0]) → COND_LOAD731(i20[0] >= 0 && i20[0] <= 1000 && i20[0] + 2 > 0, i19[0], i20[0])
(1): COND_LOAD731(TRUE, i19[1], i20[1]) → LOAD731(i19[1] + i20[1], i20[1] + 2)

(0) -> (1), if ((i19[0] →* i19[1])∧(i20[0] →* i20[1])∧(i20[0] >= 0 && i20[0] <= 1000 && i20[0] + 2 > 0 →* TRUE))


(1) -> (0), if ((i19[1] + i20[1] →* i19[0])∧(i20[1] + 2 →* i20[0]))



The set Q consists of the following terms:
Load731(x0, x1)
Cond_Load731(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): LOAD731(i19[0], i20[0]) → COND_LOAD731(i20[0] >= 0 && i20[0] <= 1000 && i20[0] + 2 > 0, i19[0], i20[0])
(1): COND_LOAD731(TRUE, i19[1], i20[1]) → LOAD731(i19[1] + i20[1], i20[1] + 2)

(0) -> (1), if ((i19[0] →* i19[1])∧(i20[0] →* i20[1])∧(i20[0] >= 0 && i20[0] <= 1000 && i20[0] + 2 > 0 →* TRUE))


(1) -> (0), if ((i19[1] + i20[1] →* i19[0])∧(i20[1] + 2 →* i20[0]))



The set Q consists of the following terms:
Load731(x0, x1)
Cond_Load731(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 LOAD731(i19, i20) → COND_LOAD731(&&(&&(>=(i20, 0), <=(i20, 1000)), >(+(i20, 2), 0)), i19, i20) the following chains were created:
  • We consider the chain LOAD731(i19[0], i20[0]) → COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0]), COND_LOAD731(TRUE, i19[1], i20[1]) → LOAD731(+(i19[1], i20[1]), +(i20[1], 2)) which results in the following constraint:

    (1)    (i19[0]=i19[1]∧i20[0]=i20[1]∧&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0))=TRUE ⇒ LOAD731(i19[0], i20[0])≥NonInfC∧LOAD731(i19[0], i20[0])≥COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])∧(UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥))



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

    (2)    (>(+(i20[0], 2), 0)=TRUE∧>=(i20[0], 0)=TRUE∧<=(i20[0], 1000)=TRUE ⇒ LOAD731(i19[0], i20[0])≥NonInfC∧LOAD731(i19[0], i20[0])≥COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])∧(UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥))



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

    (3)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥)∧[(-1)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i20[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (4)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥)∧[(-1)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i20[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (5)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥)∧[(-1)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i20[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (6)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥)∧0 = 0∧[(-1)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i20[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)







For Pair COND_LOAD731(TRUE, i19, i20) → LOAD731(+(i19, i20), +(i20, 2)) the following chains were created:
  • We consider the chain LOAD731(i19[0], i20[0]) → COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0]), COND_LOAD731(TRUE, i19[1], i20[1]) → LOAD731(+(i19[1], i20[1]), +(i20[1], 2)), LOAD731(i19[0], i20[0]) → COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0]) which results in the following constraint:

    (7)    (i19[0]=i19[1]∧i20[0]=i20[1]∧&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0))=TRUE∧+(i19[1], i20[1])=i19[0]1∧+(i20[1], 2)=i20[0]1 ⇒ COND_LOAD731(TRUE, i19[1], i20[1])≥NonInfC∧COND_LOAD731(TRUE, i19[1], i20[1])≥LOAD731(+(i19[1], i20[1]), +(i20[1], 2))∧(UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥))



    We simplified constraint (7) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:

    (8)    (>(+(i20[0], 2), 0)=TRUE∧>=(i20[0], 0)=TRUE∧<=(i20[0], 1000)=TRUE ⇒ COND_LOAD731(TRUE, i19[0], i20[0])≥NonInfC∧COND_LOAD731(TRUE, i19[0], i20[0])≥LOAD731(+(i19[0], i20[0]), +(i20[0], 2))∧(UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥))



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

    (9)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥)∧[(-1)bni_16 + (-1)Bound*bni_16] + [(-1)bni_16]i20[0] ≥ 0∧[2 + (-1)bso_17] ≥ 0)



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

    (10)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥)∧[(-1)bni_16 + (-1)Bound*bni_16] + [(-1)bni_16]i20[0] ≥ 0∧[2 + (-1)bso_17] ≥ 0)



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

    (11)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥)∧[(-1)bni_16 + (-1)Bound*bni_16] + [(-1)bni_16]i20[0] ≥ 0∧[2 + (-1)bso_17] ≥ 0)



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

    (12)    (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥)∧0 = 0∧[(-1)bni_16 + (-1)Bound*bni_16] + [(-1)bni_16]i20[0] ≥ 0∧0 = 0∧[2 + (-1)bso_17] ≥ 0)







To summarize, we get the following constraints P≥ for the following pairs.
  • LOAD731(i19, i20) → COND_LOAD731(&&(&&(>=(i20, 0), <=(i20, 1000)), >(+(i20, 2), 0)), i19, i20)
    • (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])), ≥)∧0 = 0∧[(-1)bni_14 + (-1)Bound*bni_14] + [(-1)bni_14]i20[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)

  • COND_LOAD731(TRUE, i19, i20) → LOAD731(+(i19, i20), +(i20, 2))
    • (i20[0] + [1] ≥ 0∧i20[0] ≥ 0∧[1000] + [-1]i20[0] ≥ 0 ⇒ (UIncreasing(LOAD731(+(i19[1], i20[1]), +(i20[1], 2))), ≥)∧0 = 0∧[(-1)bni_16 + (-1)Bound*bni_16] + [(-1)bni_16]i20[0] ≥ 0∧0 = 0∧[2 + (-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) = [2]   
POL(LOAD731(x1, x2)) = [-1] + [-1]x2   
POL(COND_LOAD731(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>:

COND_LOAD731(TRUE, i19[1], i20[1]) → LOAD731(+(i19[1], i20[1]), +(i20[1], 2))

The following pairs are in Pbound:

LOAD731(i19[0], i20[0]) → COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])
COND_LOAD731(TRUE, i19[1], i20[1]) → LOAD731(+(i19[1], i20[1]), +(i20[1], 2))

The following pairs are in P≥:

LOAD731(i19[0], i20[0]) → COND_LOAD731(&&(&&(>=(i20[0], 0), <=(i20[0], 1000)), >(+(i20[0], 2), 0)), i19[0], i20[0])

At least the following rules have been oriented under context sensitive arithmetic replacement:

TRUE1 → &&(TRUE, TRUE)1
FALSE1 → &&(TRUE, FALSE)1
FALSE1 → &&(FALSE, TRUE)1
FALSE1 → &&(FALSE, FALSE)1

(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:

Boolean, Integer


R is empty.

The integer pair graph contains the following rules and edges:
(0): LOAD731(i19[0], i20[0]) → COND_LOAD731(i20[0] >= 0 && i20[0] <= 1000 && i20[0] + 2 > 0, i19[0], i20[0])


The set Q consists of the following terms:
Load731(x0, x1)
Cond_Load731(TRUE, x0, x1)

(30) IDependencyGraphProof (EQUIVALENT transformation)

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

(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:
none


R is empty.

The integer pair graph is empty.

The set Q consists of the following terms:
Load731(x0, x1)
Cond_Load731(TRUE, x0, x1)

(33) IDependencyGraphProof (EQUIVALENT transformation)

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

(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:
Load404(1000, 1000, 1000, i4, i5) → Cond_Load404(i5 >= 0 && i5 <= 1000 && i5 + 1 > 0, 1000, 1000, 1000, i4, i5)
Cond_Load404(TRUE, 1000, 1000, 1000, i4, i5) → Load404(1000, 1000, 1000, i4 + i5, i5 + 1)
The set Q consists of the following terms:
Load404(1000, 1000, 1000, x0, x1)
Cond_Load404(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:

Load404(x1, x2, x3, x4, x5) → Load404(x4, x5)
Cond_Load404(x1, x2, x3, x4, x5, x6) → Cond_Load404(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:
Load404(i4, i5) → Cond_Load404(i5 >= 0 && i5 <= 1000 && i5 + 1 > 0, i4, i5)
Cond_Load404(TRUE, i4, i5) → Load404(i4 + i5, i5 + 1)
The set Q consists of the following terms:
Load404(x0, x1)
Cond_Load404(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:
Load404(i4, i5) → Cond_Load404(i5 >= 0 && i5 <= 1000 && i5 + 1 > 0, i4, i5)
Cond_Load404(TRUE, i4, i5) → Load404(i4 + i5, i5 + 1)

The integer pair graph contains the following rules and edges:
(0): LOAD404(i4[0], i5[0]) → COND_LOAD404(i5[0] >= 0 && i5[0] <= 1000 && i5[0] + 1 > 0, i4[0], i5[0])
(1): COND_LOAD404(TRUE, i4[1], i5[1]) → LOAD404(i4[1] + i5[1], i5[1] + 1)

(0) -> (1), if ((i5[0] >= 0 && i5[0] <= 1000 && i5[0] + 1 > 0 →* TRUE)∧(i4[0] →* i4[1])∧(i5[0] →* i5[1]))


(1) -> (0), if ((i5[1] + 1 →* i5[0])∧(i4[1] + i5[1] →* i4[0]))



The set Q consists of the following terms:
Load404(x0, x1)
Cond_Load404(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): LOAD404(i4[0], i5[0]) → COND_LOAD404(i5[0] >= 0 && i5[0] <= 1000 && i5[0] + 1 > 0, i4[0], i5[0])
(1): COND_LOAD404(TRUE, i4[1], i5[1]) → LOAD404(i4[1] + i5[1], i5[1] + 1)

(0) -> (1), if ((i5[0] >= 0 && i5[0] <= 1000 && i5[0] + 1 > 0 →* TRUE)∧(i4[0] →* i4[1])∧(i5[0] →* i5[1]))


(1) -> (0), if ((i5[1] + 1 →* i5[0])∧(i4[1] + i5[1] →* i4[0]))



The set Q consists of the following terms:
Load404(x0, x1)
Cond_Load404(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 LOAD404(i4, i5) → COND_LOAD404(&&(&&(>=(i5, 0), <=(i5, 1000)), >(+(i5, 1), 0)), i4, i5) the following chains were created:
  • We consider the chain LOAD404(i4[0], i5[0]) → COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0]), COND_LOAD404(TRUE, i4[1], i5[1]) → LOAD404(+(i4[1], i5[1]), +(i5[1], 1)) which results in the following constraint:

    (1)    (&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0))=TRUE∧i4[0]=i4[1]∧i5[0]=i5[1] ⇒ LOAD404(i4[0], i5[0])≥NonInfC∧LOAD404(i4[0], i5[0])≥COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])∧(UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥))



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

    (2)    (>(+(i5[0], 1), 0)=TRUE∧>=(i5[0], 0)=TRUE∧<=(i5[0], 1000)=TRUE ⇒ LOAD404(i4[0], i5[0])≥NonInfC∧LOAD404(i4[0], i5[0])≥COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])∧(UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥))



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

    (3)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i5[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (4)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i5[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (5)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥)∧[(-1)Bound*bni_14] + [(-1)bni_14]i5[0] ≥ 0∧[(-1)bso_15] ≥ 0)



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

    (6)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥)∧0 = 0∧[(-1)Bound*bni_14] + [(-1)bni_14]i5[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)







For Pair COND_LOAD404(TRUE, i4, i5) → LOAD404(+(i4, i5), +(i5, 1)) the following chains were created:
  • We consider the chain LOAD404(i4[0], i5[0]) → COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0]), COND_LOAD404(TRUE, i4[1], i5[1]) → LOAD404(+(i4[1], i5[1]), +(i5[1], 1)), LOAD404(i4[0], i5[0]) → COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0]) which results in the following constraint:

    (7)    (&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0))=TRUE∧i4[0]=i4[1]∧i5[0]=i5[1]∧+(i5[1], 1)=i5[0]1∧+(i4[1], i5[1])=i4[0]1 ⇒ COND_LOAD404(TRUE, i4[1], i5[1])≥NonInfC∧COND_LOAD404(TRUE, i4[1], i5[1])≥LOAD404(+(i4[1], i5[1]), +(i5[1], 1))∧(UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥))



    We simplified constraint (7) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:

    (8)    (>(+(i5[0], 1), 0)=TRUE∧>=(i5[0], 0)=TRUE∧<=(i5[0], 1000)=TRUE ⇒ COND_LOAD404(TRUE, i4[0], i5[0])≥NonInfC∧COND_LOAD404(TRUE, i4[0], i5[0])≥LOAD404(+(i4[0], i5[0]), +(i5[0], 1))∧(UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥))



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

    (9)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i5[0] ≥ 0∧[1 + (-1)bso_17] ≥ 0)



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

    (10)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i5[0] ≥ 0∧[1 + (-1)bso_17] ≥ 0)



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

    (11)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥)∧[(-1)Bound*bni_16] + [(-1)bni_16]i5[0] ≥ 0∧[1 + (-1)bso_17] ≥ 0)



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

    (12)    (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥)∧0 = 0∧[(-1)Bound*bni_16] + [(-1)bni_16]i5[0] ≥ 0∧0 = 0∧[1 + (-1)bso_17] ≥ 0)







To summarize, we get the following constraints P≥ for the following pairs.
  • LOAD404(i4, i5) → COND_LOAD404(&&(&&(>=(i5, 0), <=(i5, 1000)), >(+(i5, 1), 0)), i4, i5)
    • (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])), ≥)∧0 = 0∧[(-1)Bound*bni_14] + [(-1)bni_14]i5[0] ≥ 0∧0 = 0∧[(-1)bso_15] ≥ 0)

  • COND_LOAD404(TRUE, i4, i5) → LOAD404(+(i4, i5), +(i5, 1))
    • (i5[0] ≥ 0∧i5[0] ≥ 0∧[1000] + [-1]i5[0] ≥ 0 ⇒ (UIncreasing(LOAD404(+(i4[1], i5[1]), +(i5[1], 1))), ≥)∧0 = 0∧[(-1)Bound*bni_16] + [(-1)bni_16]i5[0] ≥ 0∧0 = 0∧[1 + (-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) = [2]   
POL(FALSE) = [2]   
POL(LOAD404(x1, x2)) = [-1]x2   
POL(COND_LOAD404(x1, x2, x3)) = [2] + [-1]x3 + [-1]x1   
POL(&&(x1, x2)) = [2]   
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_LOAD404(TRUE, i4[1], i5[1]) → LOAD404(+(i4[1], i5[1]), +(i5[1], 1))

The following pairs are in Pbound:

LOAD404(i4[0], i5[0]) → COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])
COND_LOAD404(TRUE, i4[1], i5[1]) → LOAD404(+(i4[1], i5[1]), +(i5[1], 1))

The following pairs are in P≥:

LOAD404(i4[0], i5[0]) → COND_LOAD404(&&(&&(>=(i5[0], 0), <=(i5[0], 1000)), >(+(i5[0], 1), 0)), i4[0], i5[0])

At least the following rules have been oriented under context sensitive arithmetic replacement:

&&(TRUE, TRUE)1 ↔ TRUE1
&&(TRUE, FALSE)1 ↔ FALSE1
&&(FALSE, TRUE)1 ↔ FALSE1
&&(FALSE, FALSE)1 ↔ FALSE1

(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): LOAD404(i4[0], i5[0]) → COND_LOAD404(i5[0] >= 0 && i5[0] <= 1000 && i5[0] + 1 > 0, i4[0], i5[0])


The set Q consists of the following terms:
Load404(x0, x1)
Cond_Load404(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:
none


R is empty.

The integer pair graph is empty.

The set Q consists of the following terms:
Load404(x0, x1)
Cond_Load404(TRUE, x0, x1)

(48) IDependencyGraphProof (EQUIVALENT transformation)

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

(49) TRUE