Termination w.r.t. Q proof of /tmp/tmpxiXeo0/Ex5_7_Luc97

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 PisEmptyProof (⇔)

    ↳19 TRUE

  ↳20 QDP

    ↳21 QDPOrderProof (⇔)

    ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 QDPOrderProof (⇔)

    ↳31 QDP

    ↳32 PisEmptyProof (⇔)

    ↳33 TRUE

  ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 TRUE

  ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 PisEmptyProof (⇔)

    ↳47 TRUE

  ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 PisEmptyProof (⇔)

    ↳52 TRUE

  ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 PisEmptyProof (⇔)

    ↳59 TRUE

  ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 QDPOrderProof (⇔)

    ↳64 QDP

    ↳65 PisEmptyProof (⇔)

    ↳66 TRUE

  ↳67 QDP

    ↳68 QDPOrderProof (⇔)

    ↳69 QDP

    ↳70 QDPOrderProof (⇔)

    ↳71 QDP

    ↳72 PisEmptyProof (⇔)

    ↳73 TRUE

  ↳74 QDP

    ↳75 QDPOrderProof (⇔)

    ↳76 QDP

    ↳77 QDPOrderProof (⇔)

    ↳78 QDP

    ↳79 PisEmptyProof (⇔)

    ↳80 TRUE

  ↳81 QDP

    ↳82 QDPOrderProof (⇔)

    ↳83 QDP

    ↳84 QDPOrderProof (⇔)

    ↳85 QDP

    ↳86 PisEmptyProof (⇔)

    ↳87 TRUE

  ↳88 QDP

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 14 SCCs with 44 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(ok(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

ok₁ > FROM₁

Status:

FROM₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

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

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

ok₁ > CONS₁

Status:

CONS₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

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

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

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

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(ok(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

ok₁ > S₁

Status:

S₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

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

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

QUOTE(ok(X)) → QUOTE(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(22) Obligation:

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

QUOTE(mark(X)) → QUOTE(X)

The TRS R consists of the following rules:

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

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

QUOTE(mark(X)) → QUOTE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > QUOTE₁

Status:

QUOTE₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

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

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [LPO].
Precedence:

ok₁ > SEL1₁

Status:

SEL1₁: [1]
mark₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(29) Obligation:

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

SEL1(mark(X1), X2) → SEL1(X1, X2)

The TRS R consists of the following rules:

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

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL1(mark(X1), X2) → SEL1(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > SEL1₁

Status:

SEL1₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

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

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

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) Obligation:

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

S1(ok(X)) → S1(X)
S1(mark(X)) → S1(X)

The TRS R consists of the following rules:

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

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S1(ok(X)) → S1(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

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

S1(mark(X)) → S1(X)

The TRS R consists of the following rules:

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

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

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S1(mark(X)) → S1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > S1₁

Status:

S1₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(38) Obligation:

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

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

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

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE

(41) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL1(ok(X)) → DBL1(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

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

DBL1(mark(X)) → DBL1(X)

The TRS R consists of the following rules:

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

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

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL1(mark(X)) → DBL1(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > DBL1₁

Status:

DBL1₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(45) Obligation:

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

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

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

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

INDX₁: [1]
ok₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(50) Obligation:

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

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

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

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [LPO].
Precedence:

ok₁ > SEL₁

Status:

SEL₁: [1]
mark₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

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

SEL(mark(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

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

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

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(mark(X1), X2) → SEL(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > SEL₁

Status:

SEL₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

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

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

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

(58) PisEmptyProof (EQUIVALENT transformation)

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

(59) TRUE

(60) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBLS(ok(X)) → DBLS(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(62) Obligation:

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

DBLS(mark(X)) → DBLS(X)

The TRS R consists of the following rules:

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

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

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBLS(mark(X)) → DBLS(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > DBLS₁

Status:

DBLS₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(64) Obligation:

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

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

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

(65) PisEmptyProof (EQUIVALENT transformation)

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

(66) TRUE

(67) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(68) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(ok(X)) → DBL(X)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(69) Obligation:

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

DBL(mark(X)) → DBL(X)

The TRS R consists of the following rules:

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

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

(70) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(mark(X)) → DBL(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > DBL₁

Status:

DBL₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(71) Obligation:

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

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

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

(72) PisEmptyProof (EQUIVALENT transformation)

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

(73) TRUE

(74) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(75) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

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

The following usable rules [FROCOS05] were oriented: none

(76) Obligation:

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

PROPER(quote(X)) → PROPER(X)

The TRS R consists of the following rules:

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

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

(77) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(quote(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

quote₁ > PROPER₁

Status:

PROPER₁: [1]
quote₁: [1]

The following usable rules [FROCOS05] were oriented: none

(78) Obligation:

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

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

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

(79) PisEmptyProof (EQUIVALENT transformation)

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

(80) TRUE

(81) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(82) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

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

The following usable rules [FROCOS05] were oriented: none

(83) Obligation:

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

ACTIVE(quote(X)) → ACTIVE(X)

The TRS R consists of the following rules:

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

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

(84) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(quote(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

quote₁ > ACTIVE₁

Status:

ACTIVE₁: [1]
quote₁: [1]

The following usable rules [FROCOS05] were oriented: none

(85) Obligation:

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

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

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

(86) PisEmptyProof (EQUIVALENT transformation)

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

(87) TRUE

(88) Obligation:

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

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

The TRS R consists of the following rules:

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

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