(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(s(X), Y) → S(n__add(activate(X), Y))
ADD(s(X), Y) → ACTIVATE(X)
LEN(cons(X, Z)) → S(n__len(activate(Z)))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__len(X)) → LEN(activate(X))
ACTIVATE(n__len(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → ACTIVATE(X)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__len(X)) → LEN(activate(X))
ACTIVATE(n__len(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( ADD(x1, x2) ) = x1 + x2 + 2 |
POL( FST(x1, x2) ) = x1 + x2 + 2 |
POL( n__fst(x1, x2) ) = x1 + x2 + 2 |
POL( fst(x1, x2) ) = x1 + x2 + 2 |
POL( n__from(x1) ) = 2x1 + 1 |
POL( from(x1) ) = 2x1 + 1 |
POL( n__add(x1, x2) ) = x1 + x2 |
POL( add(x1, x2) ) = x1 + x2 |
POL( n__len(x1) ) = 2x1 + 2 |
POL( ACTIVATE(x1) ) = x1 + 2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
fst(0, Z) → nil
fst(X1, X2) → n__fst(X1, X2)
add(0, X) → X
add(X1, X2) → n__add(X1, X2)
len(nil) → 0
len(X) → n__len(X)
len(cons(X, Z)) → s(n__len(activate(Z)))
s(X) → n__s(X)
add(s(X), Y) → s(n__add(activate(X), Y))
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
LEN(cons(X, Z)) → ACTIVATE(Z)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(
x1,
x2) =
x1
s(
x1) =
x1
ACTIVATE(
x1) =
x1
n__add(
x1,
x2) =
n__add(
x1,
x2)
activate(
x1) =
activate(
x1)
n__fst(
x1,
x2) =
n__fst
fst(
x1,
x2) =
fst
n__from(
x1) =
n__from
from(
x1) =
from
n__s(
x1) =
x1
add(
x1,
x2) =
add(
x1,
x2)
n__len(
x1) =
n__len
len(
x1) =
len
cons(
x1,
x2) =
cons
nil =
nil
0 =
0
Knuth-Bendix order [KBO] with precedence:
activate1 > fst > nfst
activate1 > add2 > nadd2
activate1 > cons
activate1 > from > nfrom
activate1 > len > nlen
activate1 > len > 0
activate1 > nil
and weight map:
n__len=1
n__fst=5
fst=5
add_2=1
0=1
cons=3
n__from=4
from=4
n__add_2=1
activate_1=0
len=1
nil=3
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
add(s(X), Y) → s(n__add(activate(X), Y))
len(cons(X, Z)) → s(n__len(activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
fst(0, Z) → nil
fst(X1, X2) → n__fst(X1, X2)
add(0, X) → X
add(X1, X2) → n__add(X1, X2)
len(nil) → 0
len(X) → n__len(X)
s(X) → n__s(X)
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADD(s(X), Y) → ACTIVATE(X)
The TRS R consists of the following rules:
fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(12) TRUE