(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
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:
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
FROM(X) → FROM(s(X))
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
SEL1(0, cons(X, Z)) → QUOTE(X)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, Z)
QUOTE1(cons(X, Z)) → QUOTE(X)
QUOTE1(cons(X, Z)) → QUOTE1(Z)
QUOTE(s(X)) → QUOTE(X)
QUOTE(sel(X, Z)) → SEL1(X, Z)
QUOTE1(first(X, Z)) → FIRST1(X, Z)
UNQUOTE(s1(X)) → UNQUOTE(X)
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
UNQUOTE1(cons1(X, Z)) → UNQUOTE(X)
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
The TRS R consists of the following rules:
sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
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 5 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
UNQUOTE(s1(X)) → UNQUOTE(X)
The TRS R consists of the following rules:
sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
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:
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
The TRS R consists of the following rules:
sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
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:
SEL1(0, cons(X, Z)) → QUOTE(X)
QUOTE(s(X)) → QUOTE(X)
QUOTE(sel(X, Z)) → SEL1(X, Z)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
The TRS R consists of the following rules:
sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
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:
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, Z)
The TRS R consists of the following rules:
sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
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:
QUOTE1(cons(X, Z)) → QUOTE1(Z)
The TRS R consists of the following rules:
sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
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:
FROM(X) → FROM(s(X))
The TRS R consists of the following rules:
sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
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:
FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
The TRS R consists of the following rules:
sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
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:
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
The TRS R consists of the following rules:
sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.