(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(activate(X))
activate(n__dbls(X)) → dbls(activate(X))
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(n__indx(X1, X2)) → indx(activate(X1), X2)
activate(n__from(X)) → from(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:
DBL(s(X)) → S(n__s(n__dbl(activate(X))))
DBL(s(X)) → ACTIVATE(X)
DBLS(cons(X, Y)) → ACTIVATE(X)
DBLS(cons(X, Y)) → ACTIVATE(Y)
SEL(0, cons(X, Y)) → ACTIVATE(X)
SEL(s(X), cons(Y, Z)) → SEL(activate(X), activate(Z))
SEL(s(X), cons(Y, Z)) → ACTIVATE(X)
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
INDX(cons(X, Y), Z) → ACTIVATE(X)
INDX(cons(X, Y), Z) → ACTIVATE(Z)
INDX(cons(X, Y), Z) → ACTIVATE(Y)
FROM(X) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__dbl(X)) → DBL(activate(X))
ACTIVATE(n__dbl(X)) → ACTIVATE(X)
ACTIVATE(n__dbls(X)) → DBLS(activate(X))
ACTIVATE(n__dbls(X)) → ACTIVATE(X)
ACTIVATE(n__sel(X1, X2)) → SEL(activate(X1), activate(X2))
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__indx(X1, X2)) → INDX(activate(X1), X2)
ACTIVATE(n__indx(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__from(X)) → FROM(X)
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(activate(X))
activate(n__dbls(X)) → dbls(activate(X))
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(n__indx(X1, X2)) → indx(activate(X1), X2)
activate(n__from(X)) → from(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 2 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__dbl(X)) → DBL(activate(X))
ACTIVATE(n__dbl(X)) → ACTIVATE(X)
ACTIVATE(n__dbls(X)) → DBLS(activate(X))
DBLS(cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__dbls(X)) → ACTIVATE(X)
ACTIVATE(n__sel(X1, X2)) → SEL(activate(X1), activate(X2))
SEL(0, cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__indx(X1, X2)) → INDX(activate(X1), X2)
INDX(cons(X, Y), Z) → ACTIVATE(X)
ACTIVATE(n__indx(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__from(X)) → FROM(X)
FROM(X) → ACTIVATE(X)
INDX(cons(X, Y), Z) → ACTIVATE(Z)
INDX(cons(X, Y), Z) → ACTIVATE(Y)
SEL(s(X), cons(Y, Z)) → SEL(activate(X), activate(Z))
SEL(s(X), cons(Y, Z)) → ACTIVATE(X)
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
DBLS(cons(X, Y)) → ACTIVATE(Y)
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(n__dbl(activate(X)), n__dbls(activate(Y)))
sel(0, cons(X, Y)) → activate(X)
sel(s(X), cons(Y, Z)) → sel(activate(X), activate(Z))
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(n__sel(activate(X), activate(Z)), n__indx(activate(Y), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
s(X) → n__s(X)
dbl(X) → n__dbl(X)
dbls(X) → n__dbls(X)
sel(X1, X2) → n__sel(X1, X2)
indx(X1, X2) → n__indx(X1, X2)
from(X) → n__from(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(activate(X))
activate(n__dbls(X)) → dbls(activate(X))
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(n__indx(X1, X2)) → indx(activate(X1), X2)
activate(n__from(X)) → from(X)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.