(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y
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:
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, 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:
LT(s(x), s(y)) → LT(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → TIMES(x, y)
PLUS(s(x), y) → PLUS(x, y)
FAC(x) → LOOP(x, s(0), s(0))
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
LOOP(x, c, y) → LT(x, c)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
IF(false, x, c, y) → TIMES(y, s(c))
The TRS R consists of the following rules:
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, 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 4 SCCs with 4 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(x, y)
The TRS R consists of the following rules:
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, 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:
PLUS(s(x), y) → PLUS(x, y)
R is empty.
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, 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].
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(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:
- PLUS(s(x), y) → PLUS(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(x, y)
The TRS R consists of the following rules:
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, 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:
TIMES(s(x), y) → TIMES(x, y)
R is empty.
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, 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].
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(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:
- TIMES(s(x), y) → TIMES(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:
LT(s(x), s(y)) → LT(x, y)
The TRS R consists of the following rules:
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, 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:
LT(s(x), s(y)) → LT(x, y)
R is empty.
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, 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].
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LT(s(x), s(y)) → LT(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) 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:
- LT(s(x), s(y)) → LT(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(27) YES
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
The TRS R consists of the following rules:
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
fac(x) → loop(x, s(0), s(0))
loop(x, c, y) → if(lt(x, c), x, c, y)
if(false, x, c, y) → loop(x, s(c), times(y, s(c)))
if(true, x, c, y) → y
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)
We have to consider all minimal (P,Q,R)-chains.
(29) 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.
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)
We have to consider all minimal (P,Q,R)-chains.
(31) 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].
fac(x0)
loop(x0, x1, x2)
if(false, x0, x1, x2)
if(true, x0, x1, x2)
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(33) 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,
c,
y) →
IF(
lt(
x,
c),
x,
c,
y) the following chains were created:
- We consider the chain IF(false, x3, x4, x5) → LOOP(x3, s(x4), times(x5, s(x4))), LOOP(x6, x7, x8) → IF(lt(x6, x7), x6, x7, x8) which results in the following constraint:
| (1) (LOOP(x3, s(x4), times(x5, s(x4)))=LOOP(x6, x7, x8) ⇒ LOOP(x6, x7, x8)≥IF(lt(x6, x7), x6, x7, x8)) |
We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint:
| (2) (LOOP(x3, s(x4), x8)≥IF(lt(x3, s(x4)), x3, s(x4), x8)) |
For Pair
IF(
false,
x,
c,
y) →
LOOP(
x,
s(
c),
times(
y,
s(
c))) the following chains were created:
- We consider the chain LOOP(x9, x10, x11) → IF(lt(x9, x10), x9, x10, x11), IF(false, x12, x13, x14) → LOOP(x12, s(x13), times(x14, s(x13))) which results in the following constraint:
| (1) (IF(lt(x9, x10), x9, x10, x11)=IF(false, x12, x13, x14) ⇒ IF(false, x12, x13, x14)≥LOOP(x12, s(x13), times(x14, s(x13)))) |
We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
| (2) (lt(x9, x10)=false ⇒ IF(false, x9, x10, x11)≥LOOP(x9, s(x10), times(x11, s(x10)))) |
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x9, x10)=false which results in the following new constraints:
| (3) (false=false ⇒ IF(false, x20, 0, x11)≥LOOP(x20, s(0), times(x11, s(0)))) |
| (4) (lt(x22, x21)=false∧(∀x23:lt(x22, x21)=false ⇒ IF(false, x22, x21, x23)≥LOOP(x22, s(x21), times(x23, s(x21)))) ⇒ IF(false, s(x22), s(x21), x11)≥LOOP(s(x22), s(s(x21)), times(x11, s(s(x21))))) |
We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
| (5) (IF(false, x20, 0, x11)≥LOOP(x20, s(0), times(x11, s(0)))) |
We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x23:lt(x22, x21)=false ⇒ IF(false, x22, x21, x23)≥LOOP(x22, s(x21), times(x23, s(x21)))) with σ = [x23 / x11] which results in the following new constraint:
| (6) (IF(false, x22, x21, x11)≥LOOP(x22, s(x21), times(x11, s(x21))) ⇒ IF(false, s(x22), s(x21), x11)≥LOOP(s(x22), s(s(x21)), times(x11, s(s(x21))))) |
To summarize, we get the following constraints P
≥ for the following pairs.
- LOOP(x, c, y) → IF(lt(x, c), x, c, y)
- (LOOP(x3, s(x4), x8)≥IF(lt(x3, s(x4)), x3, s(x4), x8))
- IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
- (IF(false, x20, 0, x11)≥LOOP(x20, s(0), times(x11, s(0))))
- (IF(false, x22, x21, x11)≥LOOP(x22, s(x21), times(x11, s(x21))) ⇒ IF(false, s(x22), s(x21), x11)≥LOOP(s(x22), s(s(x21)), times(x11, s(s(x21)))))
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) = 1
POL(IF(x1, x2, x3, x4)) = -1 + x1 + x2 - x3
POL(LOOP(x1, x2, x3)) = x1 - x2
POL(c) = -2
POL(false) = 0
POL(lt(x1, x2)) = 0
POL(plus(x1, x2)) = x1 + x2
POL(s(x1)) = 1 + x1
POL(times(x1, x2)) = 1 + x2
POL(true) = 0
The following pairs are in P
>:
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
The following pairs are in P
bound:
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
The following rules are usable:
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
(34) Complex Obligation (AND)
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, c, y) → LOOP(x, s(c), times(y, s(c)))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(36) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(37) TRUE
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(x, c, y) → IF(lt(x, c), x, c, y)
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
lt(0, s(x)) → true
lt(x, 0) → false
lt(s(x), s(y)) → lt(x, y)
The set Q consists of the following terms:
lt(0, s(x0))
lt(x0, 0)
lt(s(x0), s(x1))
times(0, x0)
times(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(39) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(40) TRUE