(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
INT(x, y) → IF(le(x, y), x, y)
INT(x, y) → LE(x, y)
IF(true, x, y) → INT(s(x), y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
(8) 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.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
R is empty.
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
(10) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- LE(s(x), s(y)) → LE(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(13) TRUE
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
(15) 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.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
(17) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
int(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(19) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
INT(
x,
y) →
IF(
le(
x,
y),
x,
y) we obtained the following new rules [LPAR04]:
INT(s(z0), z1) → IF(le(s(z0), z1), s(z0), z1)
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(true, x, y) → INT(s(x), y)
INT(s(z0), z1) → IF(le(s(z0), z1), s(z0), z1)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(21) NonInfProof (EQUIVALENT transformation)
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
INT(
x,
y) →
IF(
le(
x,
y),
x,
y) the following chains were created:
- We consider the chain IF(true, x, y) → INT(s(x), y), INT(x, y) → IF(le(x, y), x, y) which results in the following constraint:
(1) (INT(s(x2), x3)=INT(x4, x5) ⇒ INT(x4, x5)≥IF(le(x4, x5), x4, x5))
We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2) (INT(s(x2), x3)≥IF(le(s(x2), x3), s(x2), x3))
For Pair
IF(
true,
x,
y) →
INT(
s(
x),
y) the following chains were created:
- We consider the chain INT(x, y) → IF(le(x, y), x, y), IF(true, x, y) → INT(s(x), y) which results in the following constraint:
(3) (IF(le(x6, x7), x6, x7)=IF(true, x8, x9) ⇒ IF(true, x8, x9)≥INT(s(x8), x9))
We simplified constraint (3) using rules (I), (II), (III) which results in the following new constraint:
(4) (le(x6, x7)=true ⇒ IF(true, x6, x7)≥INT(s(x6), x7))
We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on le(x6, x7)=true which results in the following new constraints:
(5) (true=true ⇒ IF(true, 0, x12)≥INT(s(0), x12))
(6) (le(x15, x14)=true∧(le(x15, x14)=true ⇒ IF(true, x15, x14)≥INT(s(x15), x14)) ⇒ IF(true, s(x15), s(x14))≥INT(s(s(x15)), s(x14)))
We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
(7) (IF(true, 0, x12)≥INT(s(0), x12))
We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (le(x15, x14)=true ⇒ IF(true, x15, x14)≥INT(s(x15), x14)) with σ = [ ] which results in the following new constraint:
(8) (IF(true, x15, x14)≥INT(s(x15), x14) ⇒ IF(true, s(x15), s(x14))≥INT(s(s(x15)), s(x14)))
To summarize, we get the following constraints P
≥ for the following pairs.
- INT(x, y) → IF(le(x, y), x, y)
- (INT(s(x2), x3)≥IF(le(s(x2), x3), s(x2), x3))
- IF(true, x, y) → INT(s(x), y)
- (IF(true, 0, x12)≥INT(s(0), x12))
- (IF(true, x15, x14)≥INT(s(x15), x14) ⇒ IF(true, s(x15), s(x14))≥INT(s(s(x15)), s(x14)))
The constraints for P
> respective P
bound are constructed from P
≥ where we just replace every occurence of "t ≥ s" in P
≥ by "t > s" respective "t ≥
c". Here
c stands for the fresh constant used for P
bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:
POL(0) = 0
POL(IF(x1, x2, x3)) = -1 - x1 - x2 + x3
POL(INT(x1, x2)) = -1 - x1 + x2
POL(c) = -2
POL(false) = 1
POL(le(x1, x2)) = 1
POL(s(x1)) = 1 + x1
POL(true) = 1
The following pairs are in P
>:
INT(x, y) → IF(le(x, y), x, y)
The following pairs are in P
bound:
IF(true, x, y) → INT(s(x), y)
The following rules are usable:
true → le(0, y)
false → le(s(x), 0)
le(x, y) → le(s(x), s(y))
(22) Complex Obligation (AND)
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(true, x, y) → INT(s(x), y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(24) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(25) TRUE
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INT(x, y) → IF(le(x, y), x, y)
The TRS R consists of the following rules:
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(27) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(28) TRUE