(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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:
FROM(X) → CONS(X, n__from(n__s(X)))
LENGTH(n__cons(X, Y)) → S(length1(activate(Y)))
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH(n__cons(X, Y)) → ACTIVATE(Y)
LENGTH1(X) → LENGTH(activate(X))
LENGTH1(X) → ACTIVATE(X)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
The TRS R consists of the following rules:
from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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 2 SCCs with 8 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
The TRS R consists of the following rules:
from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(X) → LENGTH(activate(X))
The TRS R consists of the following rules:
from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.