(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(dbl(s(X))) → S(s(dbl(X)))
ACTIVE(dbl(s(X))) → S(dbl(X))
ACTIVE(dbl(s(X))) → DBL(X)
ACTIVE(dbls(cons(X, Y))) → CONS(dbl(X), dbls(Y))
ACTIVE(dbls(cons(X, Y))) → DBL(X)
ACTIVE(dbls(cons(X, Y))) → DBLS(Y)
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(indx(cons(X, Y), Z)) → CONS(sel(X, Z), indx(Y, Z))
ACTIVE(indx(cons(X, Y), Z)) → SEL(X, Z)
ACTIVE(indx(cons(X, Y), Z)) → INDX(Y, Z)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(dbl1(s(X))) → S1(s1(dbl1(X)))
ACTIVE(dbl1(s(X))) → S1(dbl1(X))
ACTIVE(dbl1(s(X))) → DBL1(X)
ACTIVE(sel1(s(X), cons(Y, Z))) → SEL1(X, Z)
ACTIVE(quote(s(X))) → S1(quote(X))
ACTIVE(quote(s(X))) → QUOTE(X)
ACTIVE(quote(dbl(X))) → DBL1(X)
ACTIVE(quote(sel(X, Y))) → SEL1(X, Y)
ACTIVE(dbl(X)) → DBL(active(X))
ACTIVE(dbl(X)) → ACTIVE(X)
ACTIVE(dbls(X)) → DBLS(active(X))
ACTIVE(dbls(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(indx(X1, X2)) → INDX(active(X1), X2)
ACTIVE(indx(X1, X2)) → ACTIVE(X1)
ACTIVE(dbl1(X)) → DBL1(active(X))
ACTIVE(dbl1(X)) → ACTIVE(X)
ACTIVE(s1(X)) → S1(active(X))
ACTIVE(s1(X)) → ACTIVE(X)
ACTIVE(sel1(X1, X2)) → SEL1(active(X1), X2)
ACTIVE(sel1(X1, X2)) → ACTIVE(X1)
ACTIVE(sel1(X1, X2)) → SEL1(X1, active(X2))
ACTIVE(sel1(X1, X2)) → ACTIVE(X2)
ACTIVE(quote(X)) → QUOTE(active(X))
ACTIVE(quote(X)) → ACTIVE(X)
DBL(mark(X)) → DBL(X)
DBLS(mark(X)) → DBLS(X)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
INDX(mark(X1), X2) → INDX(X1, X2)
DBL1(mark(X)) → DBL1(X)
S1(mark(X)) → S1(X)
SEL1(mark(X1), X2) → SEL1(X1, X2)
SEL1(X1, mark(X2)) → SEL1(X1, X2)
QUOTE(mark(X)) → QUOTE(X)
PROPER(dbl(X)) → DBL(proper(X))
PROPER(dbl(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(dbls(X)) → DBLS(proper(X))
PROPER(dbls(X)) → PROPER(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(indx(X1, X2)) → INDX(proper(X1), proper(X2))
PROPER(indx(X1, X2)) → PROPER(X1)
PROPER(indx(X1, X2)) → PROPER(X2)
PROPER(from(X)) → FROM(proper(X))
PROPER(from(X)) → PROPER(X)
PROPER(dbl1(X)) → DBL1(proper(X))
PROPER(dbl1(X)) → PROPER(X)
PROPER(s1(X)) → S1(proper(X))
PROPER(s1(X)) → PROPER(X)
PROPER(sel1(X1, X2)) → SEL1(proper(X1), proper(X2))
PROPER(sel1(X1, X2)) → PROPER(X1)
PROPER(sel1(X1, X2)) → PROPER(X2)
PROPER(quote(X)) → QUOTE(proper(X))
PROPER(quote(X)) → PROPER(X)
DBL(ok(X)) → DBL(X)
S(ok(X)) → S(X)
DBLS(ok(X)) → DBLS(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
INDX(ok(X1), ok(X2)) → INDX(X1, X2)
FROM(ok(X)) → FROM(X)
DBL1(ok(X)) → DBL1(X)
S1(ok(X)) → S1(X)
SEL1(ok(X1), ok(X2)) → SEL1(X1, X2)
QUOTE(ok(X)) → QUOTE(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 14 SCCs with 44 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FROM(ok(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
ok1 > FROM1

Status:
FROM1: [1]
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(ok(X1), ok(X2)) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  CONS(x2)
ok(x1)  =  ok(x1)

Lexicographic path order with status [LPO].
Precedence:
ok1 > CONS1

Status:
CONS1: [1]
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE

(15) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(ok(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
ok1 > S1

Status:
S1: [1]
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(18) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(19) TRUE

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOTE(ok(X)) → QUOTE(X)
QUOTE(mark(X)) → QUOTE(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOTE(ok(X)) → QUOTE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOTE(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOTE(mark(X)) → QUOTE(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOTE(mark(X)) → QUOTE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > QUOTE1

Status:
QUOTE1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) TRUE

(27) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL1(X1, mark(X2)) → SEL1(X1, X2)
SEL1(mark(X1), X2) → SEL1(X1, X2)
SEL1(ok(X1), ok(X2)) → SEL1(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL1(X1, mark(X2)) → SEL1(X1, X2)
SEL1(ok(X1), ok(X2)) → SEL1(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL1(x1, x2)  =  SEL1(x2)
mark(x1)  =  mark(x1)
ok(x1)  =  ok(x1)

Lexicographic path order with status [LPO].
Precedence:
ok1 > SEL11

Status:
SEL11: [1]
mark1: [1]
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(29) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL1(mark(X1), X2) → SEL1(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL1(mark(X1), X2) → SEL1(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL1(x1, x2)  =  SEL1(x1)
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Precedence:
mark1 > SEL11

Status:
SEL11: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(32) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(33) TRUE

(34) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S1(ok(X)) → S1(X)
S1(mark(X)) → S1(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S1(ok(X)) → S1(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S1(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S1(mark(X)) → S1(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S1(mark(X)) → S1(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > S11

Status:
S11: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(38) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(39) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(40) TRUE

(41) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DBL1(ok(X)) → DBL1(X)
DBL1(mark(X)) → DBL1(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DBL1(ok(X)) → DBL1(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DBL1(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DBL1(mark(X)) → DBL1(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DBL1(mark(X)) → DBL1(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > DBL11

Status:
DBL11: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(45) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(46) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(47) TRUE

(48) Obligation:

Q DP problem:
The TRS P consists of the following rules:

INDX(ok(X1), ok(X2)) → INDX(X1, X2)
INDX(mark(X1), X2) → INDX(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


INDX(ok(X1), ok(X2)) → INDX(X1, X2)
INDX(mark(X1), X2) → INDX(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INDX(x1, x2)  =  INDX(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
INDX1: [1]
ok1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(50) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(51) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(52) TRUE

(53) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x2)
mark(x1)  =  mark(x1)
ok(x1)  =  ok(x1)

Lexicographic path order with status [LPO].
Precedence:
ok1 > SEL1

Status:
SEL1: [1]
mark1: [1]
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SEL(mark(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL(mark(X1), X2) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x1)
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Precedence:
mark1 > SEL1

Status:
SEL1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(58) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(59) TRUE

(60) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DBLS(ok(X)) → DBLS(X)
DBLS(mark(X)) → DBLS(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DBLS(ok(X)) → DBLS(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DBLS(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(62) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DBLS(mark(X)) → DBLS(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DBLS(mark(X)) → DBLS(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > DBLS1

Status:
DBLS1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(64) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(65) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(66) TRUE

(67) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DBL(ok(X)) → DBL(X)
DBL(mark(X)) → DBL(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(68) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DBL(ok(X)) → DBL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DBL(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(69) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DBL(mark(X)) → DBL(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(70) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DBL(mark(X)) → DBL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > DBL1

Status:
DBL1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(71) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(72) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(73) TRUE

(74) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROPER(s(X)) → PROPER(X)
PROPER(dbl(X)) → PROPER(X)
PROPER(dbls(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(indx(X1, X2)) → PROPER(X1)
PROPER(indx(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(dbl1(X)) → PROPER(X)
PROPER(s1(X)) → PROPER(X)
PROPER(sel1(X1, X2)) → PROPER(X1)
PROPER(sel1(X1, X2)) → PROPER(X2)
PROPER(quote(X)) → PROPER(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(75) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(s(X)) → PROPER(X)
PROPER(dbl(X)) → PROPER(X)
PROPER(dbls(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(indx(X1, X2)) → PROPER(X1)
PROPER(indx(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(dbl1(X)) → PROPER(X)
PROPER(s1(X)) → PROPER(X)
PROPER(sel1(X1, X2)) → PROPER(X1)
PROPER(sel1(X1, X2)) → PROPER(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
s(x1)  =  s(x1)
dbl(x1)  =  dbl(x1)
dbls(x1)  =  dbls(x1)
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
indx(x1, x2)  =  indx(x1, x2)
from(x1)  =  from(x1)
dbl1(x1)  =  dbl1(x1)
s1(x1)  =  s1(x1)
sel1(x1, x2)  =  sel1(x1, x2)
quote(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
s1: [1]
dbl1: [1]
dbls1: [1]
cons2: [1,2]
sel2: [1,2]
indx2: [1,2]
from1: [1]
dbl11: [1]
s11: [1]
sel12: [1,2]

The following usable rules [FROCOS05] were oriented: none

(76) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROPER(quote(X)) → PROPER(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(77) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(quote(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
quote1 > PROPER1

Status:
PROPER1: [1]
quote1: [1]

The following usable rules [FROCOS05] were oriented: none

(78) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(79) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(80) TRUE

(81) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(dbls(X)) → ACTIVE(X)
ACTIVE(dbl(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(indx(X1, X2)) → ACTIVE(X1)
ACTIVE(dbl1(X)) → ACTIVE(X)
ACTIVE(s1(X)) → ACTIVE(X)
ACTIVE(sel1(X1, X2)) → ACTIVE(X1)
ACTIVE(sel1(X1, X2)) → ACTIVE(X2)
ACTIVE(quote(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(82) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(dbls(X)) → ACTIVE(X)
ACTIVE(dbl(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(indx(X1, X2)) → ACTIVE(X1)
ACTIVE(dbl1(X)) → ACTIVE(X)
ACTIVE(s1(X)) → ACTIVE(X)
ACTIVE(sel1(X1, X2)) → ACTIVE(X1)
ACTIVE(sel1(X1, X2)) → ACTIVE(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
dbls(x1)  =  dbls(x1)
dbl(x1)  =  dbl(x1)
sel(x1, x2)  =  sel(x1, x2)
indx(x1, x2)  =  indx(x1)
dbl1(x1)  =  dbl1(x1)
s1(x1)  =  s1(x1)
sel1(x1, x2)  =  sel1(x1, x2)
quote(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
dbls1: [1]
dbl1: [1]
sel2: [1,2]
indx1: [1]
dbl11: [1]
s11: [1]
sel12: [1,2]

The following usable rules [FROCOS05] were oriented: none

(83) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(quote(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(84) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(quote(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
quote1 > ACTIVE1

Status:
ACTIVE1: [1]
quote1: [1]

The following usable rules [FROCOS05] were oriented: none

(85) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(86) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(87) TRUE

(88) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl1(0)) → mark(01)
active(dbl1(s(X))) → mark(s1(s1(dbl1(X))))
active(sel1(0, cons(X, Y))) → mark(X)
active(sel1(s(X), cons(Y, Z))) → mark(sel1(X, Z))
active(quote(0)) → mark(01)
active(quote(s(X))) → mark(s1(quote(X)))
active(quote(dbl(X))) → mark(dbl1(X))
active(quote(sel(X, Y))) → mark(sel1(X, Y))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl1(X)) → dbl1(active(X))
active(s1(X)) → s1(active(X))
active(sel1(X1, X2)) → sel1(active(X1), X2)
active(sel1(X1, X2)) → sel1(X1, active(X2))
active(quote(X)) → quote(active(X))
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
dbl1(mark(X)) → mark(dbl1(X))
s1(mark(X)) → mark(s1(X))
sel1(mark(X1), X2) → mark(sel1(X1, X2))
sel1(X1, mark(X2)) → mark(sel1(X1, X2))
quote(mark(X)) → mark(quote(X))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(dbl1(X)) → dbl1(proper(X))
proper(01) → ok(01)
proper(s1(X)) → s1(proper(X))
proper(sel1(X1, X2)) → sel1(proper(X1), proper(X2))
proper(quote(X)) → quote(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
dbl1(ok(X)) → ok(dbl1(X))
s1(ok(X)) → ok(s1(X))
sel1(ok(X1), ok(X2)) → ok(sel1(X1, X2))
quote(ok(X)) → ok(quote(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.