(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(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)
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))
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))
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))
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(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)
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)
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)
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)
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(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)
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))
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))
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))
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 10 SCCs with 27 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(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)
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))
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))
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))
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: Combined order from the following AFS and order.
FROM(x1)  =  FROM(x1)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
dbl(x1)  =  dbl(x1)
0  =  0
mark(x1)  =  x1
s(x1)  =  s(x1)
dbls(x1)  =  dbls(x1)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  x2
indx(x1, x2)  =  indx(x2)
from(x1)  =  from(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
active1 > dbl1 > s1 > ok1
active1 > 0 > ok1
active1 > dbls1 > nil
active1 > dbls1 > cons2 > ok1
active1 > indx1 > nil
active1 > indx1 > cons2 > ok1
active1 > from1 > ok1
proper1 > dbl1 > s1 > ok1
proper1 > dbls1 > nil
proper1 > dbls1 > cons2 > ok1
proper1 > indx1 > nil
proper1 > indx1 > cons2 > ok1
proper1 > from1 > ok1

Status:
FROM1: [1]
ok1: multiset
active1: multiset
dbl1: [1]
0: multiset
s1: [1]
dbls1: multiset
nil: multiset
cons2: multiset
indx1: [1]
from1: [1]
proper1: [1]
top: multiset

The following usable rules [FROCOS05] were oriented:

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

(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(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)
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))
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))
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))
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(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)
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))
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))
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))
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(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
dbl(x1)  =  dbl(x1)
0  =  0
mark(x1)  =  mark
s(x1)  =  x1
dbls(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  x1
sel(x1, x2)  =  x2
indx(x1, x2)  =  x2
from(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
CONS1 > mark
proper1 > dbl1 > ok1 > top > mark
proper1 > 0 > ok1 > top > mark
proper1 > nil > mark

Status:
CONS1: [1]
ok1: multiset
dbl1: [1]
0: multiset
mark: multiset
nil: multiset
proper1: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(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(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)
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))
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))
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))
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(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)
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))
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))
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))
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: Combined order from the following AFS and order.
S(x1)  =  S(x1)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
dbl(x1)  =  dbl(x1)
0  =  0
mark(x1)  =  x1
s(x1)  =  s(x1)
dbls(x1)  =  dbls(x1)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  x2
indx(x1, x2)  =  indx(x2)
from(x1)  =  from(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
active1 > dbl1 > s1 > ok1
active1 > 0 > ok1
active1 > dbls1 > nil
active1 > dbls1 > cons2 > ok1
active1 > indx1 > nil
active1 > indx1 > cons2 > ok1
active1 > from1 > ok1
proper1 > dbl1 > s1 > ok1
proper1 > dbls1 > nil
proper1 > dbls1 > cons2 > ok1
proper1 > indx1 > nil
proper1 > indx1 > cons2 > ok1
proper1 > from1 > ok1

Status:
S1: [1]
ok1: multiset
active1: multiset
dbl1: [1]
0: multiset
s1: [1]
dbls1: multiset
nil: multiset
cons2: multiset
indx1: [1]
from1: [1]
proper1: [1]
top: multiset

The following usable rules [FROCOS05] were oriented:

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

(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(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)
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))
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))
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))
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:

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(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)
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))
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))
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))
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.


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)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
dbl(x1)  =  dbl(x1)
0  =  0
s(x1)  =  x1
dbls(x1)  =  dbls(x1)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
indx(x1, x2)  =  indx(x1, x2)
from(x1)  =  from(x1)
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
INDX1 > from1
active1 > dbl1 > mark1 > top > from1
active1 > dbls1 > cons2 > mark1 > top > from1
active1 > sel2 > mark1 > top > from1
active1 > indx2 > mark1 > top > from1
0 > mark1 > top > from1
nil > mark1 > top > from1
proper1 > dbl1 > mark1 > top > from1
proper1 > dbls1 > cons2 > mark1 > top > from1
proper1 > sel2 > mark1 > top > from1
proper1 > indx2 > mark1 > top > from1

Status:
INDX1: multiset
mark1: multiset
active1: [1]
dbl1: multiset
0: multiset
dbls1: multiset
nil: multiset
cons2: [2,1]
sel2: multiset
indx2: [2,1]
from1: multiset
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(22) Obligation:

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

INDX(ok(X1), ok(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(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)
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))
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))
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))
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.


INDX(ok(X1), ok(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)
active(x1)  =  x1
dbl(x1)  =  dbl(x1)
0  =  0
mark(x1)  =  mark
s(x1)  =  x1
dbls(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  x1
sel(x1, x2)  =  x2
indx(x1, x2)  =  x2
from(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
INDX1 > mark
proper1 > dbl1 > ok1 > top > mark
proper1 > 0 > ok1 > top > mark
proper1 > nil > mark

Status:
INDX1: [1]
ok1: multiset
dbl1: [1]
0: multiset
mark: multiset
nil: multiset
proper1: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(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(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)
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))
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))
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))
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:

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(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)
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))
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))
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))
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.


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)
ok(x1)  =  x1
active(x1)  =  active(x1)
dbl(x1)  =  dbl(x1)
0  =  0
s(x1)  =  x1
dbls(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
indx(x1, x2)  =  indx(x1, x2)
from(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
top > active1 > dbl1 > mark1 > SEL1
top > active1 > 0 > mark1 > SEL1
top > active1 > sel2 > mark1 > SEL1
top > active1 > indx2 > mark1 > SEL1
top > active1 > indx2 > nil > SEL1
top > active1 > indx2 > cons2 > SEL1
top > proper1 > dbl1 > mark1 > SEL1
top > proper1 > sel2 > mark1 > SEL1
top > proper1 > indx2 > mark1 > SEL1
top > proper1 > indx2 > nil > SEL1
top > proper1 > indx2 > cons2 > SEL1

Status:
SEL1: multiset
mark1: multiset
active1: [1]
dbl1: [1]
0: multiset
nil: multiset
cons2: multiset
sel2: [2,1]
indx2: [2,1]
proper1: [1]
top: multiset

The following usable rules [FROCOS05] were oriented:

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

(29) Obligation:

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

SEL(X1, mark(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(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)
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))
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))
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))
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.


SEL(X1, mark(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)  =  x1
active(x1)  =  active(x1)
dbl(x1)  =  x1
0  =  0
s(x1)  =  x1
dbls(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
indx(x1, x2)  =  indx(x1, x2)
from(x1)  =  from(x1)
proper(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
SEL1 > nil
active1 > 0 > nil
active1 > cons2 > mark1 > nil
active1 > sel2 > mark1 > nil
active1 > indx2 > mark1 > nil
active1 > from1 > mark1 > nil
top > nil

Status:
SEL1: [1]
mark1: multiset
active1: multiset
0: multiset
nil: multiset
cons2: multiset
sel2: [1,2]
indx2: multiset
from1: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(31) Obligation:

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

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(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)
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))
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))
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))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
dbl(x1)  =  dbl(x1)
0  =  0
mark(x1)  =  mark
s(x1)  =  x1
dbls(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  x1
sel(x1, x2)  =  x2
indx(x1, x2)  =  x2
from(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
SEL1 > mark
proper1 > dbl1 > ok1 > top > mark
proper1 > 0 > ok1 > top > mark
proper1 > nil > mark

Status:
SEL1: [1]
ok1: multiset
dbl1: [1]
0: multiset
mark: multiset
nil: multiset
proper1: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(33) 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(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)
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))
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))
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))
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.

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) 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(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)
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))
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))
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))
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.


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
active(x1)  =  active(x1)
dbl(x1)  =  x1
0  =  0
s(x1)  =  x1
dbls(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  sel(x2)
indx(x1, x2)  =  x2
from(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
top > active1 > 0 > nil
top > active1 > cons2 > ok1 > nil
top > active1 > sel1 > ok1 > nil
top > proper1 > 0 > nil
top > proper1 > cons2 > ok1 > nil
top > proper1 > sel1 > ok1 > nil

Status:
ok1: [1]
active1: [1]
0: multiset
nil: multiset
cons2: [1,2]
sel1: [1]
proper1: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(38) 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(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)
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))
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))
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))
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) 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: Combined order from the following AFS and order.
DBLS(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
dbl(x1)  =  dbl(x1)
0  =  0
s(x1)  =  x1
dbls(x1)  =  dbls(x1)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
indx(x1, x2)  =  indx(x1, x2)
from(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
active1 > 0 > mark1
active1 > 0 > ok > top
active1 > dbls1 > dbl1 > mark1
active1 > dbls1 > dbl1 > ok > top
active1 > dbls1 > cons2 > mark1
active1 > sel2 > mark1
active1 > indx2 > mark1
nil > mark1
nil > ok > top

Status:
mark1: multiset
active1: [1]
dbl1: multiset
0: multiset
dbls1: [1]
nil: multiset
cons2: multiset
sel2: multiset
indx2: multiset
ok: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(40) 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(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)
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))
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))
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))
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.

(41) PisEmptyProof (EQUIVALENT transformation)

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

(42) TRUE

(43) 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(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)
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))
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))
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))
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.


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
active(x1)  =  active(x1)
dbl(x1)  =  x1
0  =  0
s(x1)  =  x1
dbls(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  sel(x2)
indx(x1, x2)  =  x2
from(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
top > active1 > 0 > nil
top > active1 > cons2 > ok1 > nil
top > active1 > sel1 > ok1 > nil
top > proper1 > 0 > nil
top > proper1 > cons2 > ok1 > nil
top > proper1 > sel1 > ok1 > nil

Status:
ok1: [1]
active1: [1]
0: multiset
nil: multiset
cons2: [1,2]
sel1: [1]
proper1: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(45) 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(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)
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))
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))
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))
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) 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: Combined order from the following AFS and order.
DBL(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
dbl(x1)  =  dbl(x1)
0  =  0
s(x1)  =  x1
dbls(x1)  =  dbls(x1)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
indx(x1, x2)  =  indx(x1, x2)
from(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
active1 > 0 > mark1
active1 > 0 > ok > top
active1 > dbls1 > dbl1 > mark1
active1 > dbls1 > dbl1 > ok > top
active1 > dbls1 > cons2 > mark1
active1 > sel2 > mark1
active1 > indx2 > mark1
nil > mark1
nil > ok > top

Status:
mark1: multiset
active1: [1]
dbl1: multiset
0: multiset
dbls1: [1]
nil: multiset
cons2: multiset
sel2: multiset
indx2: multiset
ok: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(47) 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(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)
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))
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))
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))
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.

(48) PisEmptyProof (EQUIVALENT transformation)

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

(49) TRUE

(50) 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)

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(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)
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))
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))
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))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(dbl(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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
s(x1)  =  x1
dbl(x1)  =  dbl(x1)
dbls(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
indx(x1, x2)  =  indx(x1, x2)
from(x1)  =  from(x1)
active(x1)  =  active(x1)
0  =  0
mark(x1)  =  x1
nil  =  nil
proper(x1)  =  proper(x1)
ok(x1)  =  ok(x1)
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
active1 > dbl1 > 0 > PROPER1
active1 > dbl1 > ok1 > top > PROPER1
active1 > indx2 > cons2 > ok1 > top > PROPER1
active1 > indx2 > sel2 > ok1 > top > PROPER1
active1 > indx2 > nil > PROPER1
active1 > from1 > cons2 > ok1 > top > PROPER1
proper1 > dbl1 > 0 > PROPER1
proper1 > dbl1 > ok1 > top > PROPER1
proper1 > indx2 > cons2 > ok1 > top > PROPER1
proper1 > indx2 > sel2 > ok1 > top > PROPER1
proper1 > indx2 > nil > PROPER1
proper1 > from1 > cons2 > ok1 > top > PROPER1

Status:
PROPER1: multiset
dbl1: [1]
cons2: multiset
sel2: [2,1]
indx2: multiset
from1: [1]
active1: multiset
0: multiset
nil: multiset
proper1: [1]
ok1: [1]
top: multiset

The following usable rules [FROCOS05] were oriented:

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

(52) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(dbls(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(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)
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))
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))
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))
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.

(53) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
s(x1)  =  s(x1)
dbls(x1)  =  x1
active(x1)  =  active(x1)
dbl(x1)  =  x1
0  =  0
mark(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  x2
indx(x1, x2)  =  indx(x1, x2)
from(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
PROPER1 > s1
top > active1 > 0 > s1
top > active1 > nil > s1
top > active1 > cons2 > indx2 > s1
top > proper1 > 0 > s1
top > proper1 > nil > s1
top > proper1 > cons2 > indx2 > s1

Status:
PROPER1: [1]
s1: multiset
active1: [1]
0: multiset
nil: multiset
cons2: multiset
indx2: [2,1]
proper1: [1]
top: multiset

The following usable rules [FROCOS05] were oriented:

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

(54) Obligation:

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

PROPER(dbls(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(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)
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))
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))
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))
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.

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(dbls(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
dbls(x1)  =  dbls(x1)
active(x1)  =  active(x1)
dbl(x1)  =  dbl
0  =  0
mark(x1)  =  mark
s(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons
sel(x1, x2)  =  sel
indx(x1, x2)  =  indx
from(x1)  =  from
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
PROPER1 > dbls1
dbl > active1 > mark > proper1 > 0 > dbls1
dbl > active1 > nil > dbls1
dbl > active1 > cons > proper1 > 0 > dbls1
indx > sel > active1 > mark > proper1 > 0 > dbls1
indx > sel > active1 > nil > dbls1
indx > sel > active1 > cons > proper1 > 0 > dbls1
from > mark > proper1 > 0 > dbls1
from > cons > proper1 > 0 > dbls1
top > proper1 > 0 > dbls1

Status:
PROPER1: multiset
dbls1: multiset
active1: multiset
dbl: []
0: multiset
mark: multiset
nil: multiset
cons: []
sel: multiset
indx: []
from: multiset
proper1: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(56) 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(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)
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))
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))
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))
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.

(57) PisEmptyProof (EQUIVALENT transformation)

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

(58) TRUE

(59) 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)

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(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)
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))
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))
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))
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.

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(indx(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
dbls(x1)  =  x1
dbl(x1)  =  x1
sel(x1, x2)  =  sel(x1, x2)
indx(x1, x2)  =  indx(x1, x2)
active(x1)  =  active(x1)
0  =  0
mark(x1)  =  mark
s(x1)  =  s
nil  =  nil
cons(x1, x2)  =  x1
from(x1)  =  from
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
ACTIVE1 > mark
active1 > indx2 > ok > sel2 > mark
active1 > 0 > ok > sel2 > mark
active1 > nil > ok > sel2 > mark
active1 > from > ok > sel2 > mark
s > ok > sel2 > mark
top > mark

Status:
ACTIVE1: [1]
sel2: [1,2]
indx2: multiset
active1: [1]
0: multiset
mark: multiset
s: multiset
nil: multiset
from: []
ok: multiset
top: multiset

The following usable rules [FROCOS05] were oriented:

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

(61) Obligation:

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

ACTIVE(dbls(X)) → ACTIVE(X)
ACTIVE(dbl(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(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)
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))
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))
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))
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.

(62) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
dbls(x1)  =  dbls(x1)
dbl(x1)  =  x1
active(x1)  =  x1
0  =  0
mark(x1)  =  mark
s(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  x1
sel(x1, x2)  =  x2
indx(x1, x2)  =  x2
from(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
top > proper1 > dbls1 > ACTIVE1 > mark
top > proper1 > dbls1 > nil > mark
top > proper1 > 0 > mark

Status:
ACTIVE1: multiset
dbls1: [1]
0: multiset
mark: []
nil: multiset
proper1: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(63) Obligation:

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

ACTIVE(dbl(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(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)
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))
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))
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))
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.

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(dbl(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
dbl(x1)  =  dbl(x1)
active(x1)  =  active(x1)
0  =  0
mark(x1)  =  x1
s(x1)  =  s
dbls(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sel(x1, x2)  =  sel(x2)
indx(x1, x2)  =  indx(x2)
from(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive path order with status [RPO].
Precedence:
active1 > dbl1 > 0
active1 > dbl1 > s
active1 > nil
active1 > cons2 > indx1
active1 > sel1

Status:
ACTIVE1: [1]
dbl1: [1]
active1: [1]
0: multiset
s: multiset
nil: multiset
cons2: multiset
sel1: multiset
indx1: multiset
top: []

The following usable rules [FROCOS05] were oriented:

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

(65) 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(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)
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))
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))
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))
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.

(66) PisEmptyProof (EQUIVALENT transformation)

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

(67) TRUE

(68) 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(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)
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))
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))
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))
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.