(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
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)) → DBL(X)
DBLS(cons(X, Y)) → DBL(X)
DBLS(cons(X, Y)) → DBLS(Y)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
INDX(cons(X, Y), Z) → SEL(X, Z)
INDX(cons(X, Y), Z) → INDX(Y, Z)
FROM(X) → FROM(s(X))
DBL1(s(X)) → DBL1(X)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
QUOTE(s(X)) → QUOTE(X)
QUOTE(dbl(X)) → DBL1(X)
QUOTE(sel(X, Y)) → SEL1(X, Y)
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
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 8 SCCs with 4 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
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:
DBL1(s(X)) → DBL1(X)
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
QUOTE(s(X)) → QUOTE(X)
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(X) → FROM(s(X))
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INDX(cons(X, Y), Z) → INDX(Y, Z)
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DBL(s(X)) → DBL(X)
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DBLS(cons(X, Y)) → DBLS(Y)
The TRS R consists of the following rules:
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.