(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))
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(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
(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)
DOUBLE(s(x)) → DOUBLE(x)
LOG(s(x)) → LOOP(s(x), s(0), 0)
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)
LOOP(x, s(y), z) → LE(x, s(y))
IF(false, x, y, z) → LOOP(x, double(y), s(z))
IF(false, x, y, z) → DOUBLE(y)
The TRS R consists of the following rules:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DOUBLE(s(x)) → DOUBLE(x)
The TRS R consists of the following rules:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
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:
DOUBLE(s(x)) → DOUBLE(x)
R is empty.
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
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(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DOUBLE(s(x)) → DOUBLE(x)
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:
- DOUBLE(s(x)) → DOUBLE(x)
The graph contains the following edges 1 > 1
(13) YES
(14) 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(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
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:
LE(s(x), s(y)) → LE(x, y)
R is empty.
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
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].
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
(18) 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.
(19) 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
(20) YES
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)
IF(false, x, y, z) → LOOP(x, double(y), s(z))
The TRS R consists of the following rules:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
We have to consider all minimal (P,Q,R)-chains.
(22) 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.
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)
IF(false, x, y, z) → LOOP(x, double(y), s(z))
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
We have to consider all minimal (P,Q,R)-chains.
(24) 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].
log(0)
log(s(x0))
loop(x0, s(x1), x2)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)
IF(false, x, y, z) → LOOP(x, double(y), s(z))
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(26) 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
LOOP(
x,
s(
y),
z) →
IF(
le(
x,
s(
y)),
x,
s(
y),
z) the following chains were created:
- We consider the chain LOOP(x3, s(x4), x5) → IF(le(x3, s(x4)), x3, s(x4), x5), IF(false, x6, x7, x8) → LOOP(x6, double(x7), s(x8)) which results in the following constraint:
| (1) (IF(le(x3, s(x4)), x3, s(x4), x5)=IF(false, x6, x7, x8) ⇒ LOOP(x3, s(x4), x5)≥IF(le(x3, s(x4)), x3, s(x4), x5)) |
We simplified constraint (1) using rules (I), (II), (IV), (VII) which results in the following new constraint:
| (2) (s(x4)=x18∧le(x3, x18)=false ⇒ LOOP(x3, s(x4), x5)≥IF(le(x3, s(x4)), x3, s(x4), x5)) |
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x3, x18)=false which results in the following new constraints:
| (3) (le(x21, x20)=false∧s(x4)=s(x20)∧(∀x22,x23:le(x21, x20)=false∧s(x22)=x20 ⇒ LOOP(x21, s(x22), x23)≥IF(le(x21, s(x22)), x21, s(x22), x23)) ⇒ LOOP(s(x21), s(x4), x5)≥IF(le(s(x21), s(x4)), s(x21), s(x4), x5)) |
| (4) (false=false∧s(x4)=0 ⇒ LOOP(s(x24), s(x4), x5)≥IF(le(s(x24), s(x4)), s(x24), s(x4), x5)) |
We simplified constraint (3) using rules (I), (II), (III), (IV) which results in the following new constraint:
| (5) (le(x21, x20)=false ⇒ LOOP(s(x21), s(x20), x5)≥IF(le(s(x21), s(x20)), s(x21), s(x20), x5)) |
We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on le(x21, x20)=false which results in the following new constraints:
| (6) (le(x27, x26)=false∧(∀x28:le(x27, x26)=false ⇒ LOOP(s(x27), s(x26), x28)≥IF(le(s(x27), s(x26)), s(x27), s(x26), x28)) ⇒ LOOP(s(s(x27)), s(s(x26)), x5)≥IF(le(s(s(x27)), s(s(x26))), s(s(x27)), s(s(x26)), x5)) |
| (7) (false=false ⇒ LOOP(s(s(x29)), s(0), x5)≥IF(le(s(s(x29)), s(0)), s(s(x29)), s(0), x5)) |
We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (∀x28:le(x27, x26)=false ⇒ LOOP(s(x27), s(x26), x28)≥IF(le(s(x27), s(x26)), s(x27), s(x26), x28)) with σ = [x28 / x5] which results in the following new constraint:
| (8) (LOOP(s(x27), s(x26), x5)≥IF(le(s(x27), s(x26)), s(x27), s(x26), x5) ⇒ LOOP(s(s(x27)), s(s(x26)), x5)≥IF(le(s(s(x27)), s(s(x26))), s(s(x27)), s(s(x26)), x5)) |
We simplified constraint (7) using rules (I), (II) which results in the following new constraint:
| (9) (LOOP(s(s(x29)), s(0), x5)≥IF(le(s(s(x29)), s(0)), s(s(x29)), s(0), x5)) |
For Pair
IF(
false,
x,
y,
z) →
LOOP(
x,
double(
y),
s(
z)) the following chains were created:
- We consider the chain IF(false, x9, x10, x11) → LOOP(x9, double(x10), s(x11)), LOOP(x12, s(x13), x14) → IF(le(x12, s(x13)), x12, s(x13), x14) which results in the following constraint:
| (1) (LOOP(x9, double(x10), s(x11))=LOOP(x12, s(x13), x14) ⇒ IF(false, x9, x10, x11)≥LOOP(x9, double(x10), s(x11))) |
We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
| (2) (double(x10)=s(x13) ⇒ IF(false, x9, x10, x11)≥LOOP(x9, double(x10), s(x11))) |
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on double(x10)=s(x13) which results in the following new constraint:
| (3) (s(s(double(x30)))=s(x13)∧(∀x31,x32,x33:double(x30)=s(x31) ⇒ IF(false, x32, x30, x33)≥LOOP(x32, double(x30), s(x33))) ⇒ IF(false, x9, s(x30), x11)≥LOOP(x9, double(s(x30)), s(x11))) |
We simplified constraint (3) using rules (I), (II), (IV) which results in the following new constraint:
| (4) (IF(false, x9, s(x30), x11)≥LOOP(x9, double(s(x30)), s(x11))) |
To summarize, we get the following constraints P
≥ for the following pairs.
- LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)
- (LOOP(s(x27), s(x26), x5)≥IF(le(s(x27), s(x26)), s(x27), s(x26), x5) ⇒ LOOP(s(s(x27)), s(s(x26)), x5)≥IF(le(s(s(x27)), s(s(x26))), s(s(x27)), s(s(x26)), x5))
- (LOOP(s(s(x29)), s(0), x5)≥IF(le(s(s(x29)), s(0)), s(s(x29)), s(0), x5))
- IF(false, x, y, z) → LOOP(x, double(y), s(z))
- (IF(false, x9, s(x30), x11)≥LOOP(x9, double(s(x30)), s(x11)))
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, x4)) = -1 - x1 + x2 - x3
POL(LOOP(x1, x2, x3)) = -1 + x1 - x2
POL(c) = -2
POL(double(x1)) = 2·x1
POL(false) = 0
POL(le(x1, x2)) = 0
POL(s(x1)) = 1 + x1
POL(true) = 0
The following pairs are in P
>:
IF(false, x, y, z) → LOOP(x, double(y), s(z))
The following pairs are in P
bound:
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)
The following rules are usable:
true → le(0, y)
le(x, y) → le(s(x), s(y))
0 → double(0)
s(s(double(x))) → double(s(x))
false → le(s(x), 0)
(27) Complex Obligation (AND)
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(x, s(y), z) → IF(le(x, s(y)), x, s(y), z)
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(29) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(30) TRUE
(31) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, y, z) → LOOP(x, double(y), s(z))
The TRS R consists of the following rules:
double(0) → 0
double(s(x)) → s(s(double(x)))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
double(0)
double(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(32) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(33) TRUE