Termination w.r.t. Q proof of /tmp/tmpFAsU3b/Ex3_2_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 QDPOrderProof (⇔)

    ↳33 QDP

    ↳34 PisEmptyProof (⇔)

    ↳35 TRUE

  ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 QDPOrderProof (⇔)

    ↳40 QDP

    ↳41 PisEmptyProof (⇔)

    ↳42 TRUE

  ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 QDPOrderProof (⇔)

    ↳47 QDP

    ↳48 PisEmptyProof (⇔)

    ↳49 TRUE

  ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

    ↳53 QDPOrderProof (⇔)

    ↳54 QDP

    ↳55 QDPOrderProof (⇔)

    ↳56 QDP

    ↳57 PisEmptyProof (⇔)

    ↳58 TRUE

  ↳59 QDP

    ↳60 QDPOrderProof (⇔)

    ↳61 QDP

    ↳62 QDPOrderProof (⇔)

    ↳63 QDP

    ↳64 PisEmptyProof (⇔)

    ↳65 TRUE

  ↳66 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(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

ACTIVE(dbl(s(X))) → S(s(dbl(X)))
ACTIVE(dbl(s(X))) → S(dbl(X))
ACTIVE(dbl(s(X))) → DBL(X)
ACTIVE(dbls(cons(X, Y))) → CONS(dbl(X), dbls(Y))
ACTIVE(dbls(cons(X, Y))) → DBL(X)
ACTIVE(dbls(cons(X, Y))) → DBLS(Y)
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(indx(cons(X, Y), Z)) → CONS(sel(X, Z), indx(Y, Z))
ACTIVE(indx(cons(X, Y), Z)) → SEL(X, Z)
ACTIVE(indx(cons(X, Y), Z)) → INDX(Y, Z)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(dbl(X)) → DBL(active(X))
ACTIVE(dbl(X)) → ACTIVE(X)
ACTIVE(dbls(X)) → DBLS(active(X))
ACTIVE(dbls(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(indx(X1, X2)) → INDX(active(X1), X2)
ACTIVE(indx(X1, X2)) → ACTIVE(X1)
DBL(mark(X)) → DBL(X)
DBLS(mark(X)) → DBLS(X)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
INDX(mark(X1), X2) → INDX(X1, X2)
PROPER(dbl(X)) → DBL(proper(X))
PROPER(dbl(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(dbls(X)) → DBLS(proper(X))
PROPER(dbls(X)) → PROPER(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(indx(X1, X2)) → INDX(proper(X1), proper(X2))
PROPER(indx(X1, X2)) → PROPER(X1)
PROPER(indx(X1, X2)) → PROPER(X2)
PROPER(from(X)) → FROM(proper(X))
PROPER(from(X)) → PROPER(X)
DBL(ok(X)) → DBL(X)
S(ok(X)) → S(X)
DBLS(ok(X)) → DBLS(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
INDX(ok(X1), ok(X2)) → INDX(X1, X2)
FROM(ok(X)) → FROM(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 10 SCCs with 27 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(ok(X)) → FROM(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

proper₁ > [ok₁, 0, mark, nil, top]

Status:

ok₁: [1]
0: []
mark: []
nil: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

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

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

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

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[active₁, cons₂, indx₂, from₁, proper₁] > sel₂ > [CONS₁, ok₁, 0, nil, top]

Status:

CONS₁: [1]
ok₁: [1]
active₁: [1]
0: []
nil: []
cons₂: [1,2]
sel₂: [1,2]
indx₂: [2,1]
from₁: [1]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

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

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

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

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(ok(X)) → S(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

proper₁ > [ok₁, 0, mark, nil, top]

Status:

ok₁: [1]
0: []
mark: []
nil: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(17) Obligation:

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

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

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INDX(ok(X1), ok(X2)) → INDX(X1, X2)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

active₁ > [indx₂, from₁, proper₁] > [INDX₁, ok₁, mark, s₁, cons₁, sel₁, top]
0 > [INDX₁, ok₁, mark, s₁, cons₁, sel₁, top]
nil > [INDX₁, ok₁, mark, s₁, cons₁, sel₁, top]

Status:

INDX₁: [1]
ok₁: [1]
mark: []
active₁: [1]
0: []
s₁: [1]
nil: []
cons₁: [1]
sel₁: [1]
indx₂: [2,1]
from₁: [1]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(22) Obligation:

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

INDX(mark(X1), X2) → INDX(X1, X2)

The TRS R consists of the following rules:

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

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INDX(mark(X1), X2) → INDX(X1, X2)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

proper₁ > [active₁, indx₂] > [cons₂, sel₂] > [mark₁, 0, ok₁, top]
proper₁ > nil > [mark₁, 0, ok₁, top]

Status:

mark₁: [1]
active₁: [1]
0: []
nil: []
cons₂: [2,1]
sel₂: [2,1]
indx₂: [2,1]
proper₁: [1]
ok₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(24) Obligation:

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

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

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

active₁ > 0 > [ok₁, dbl₁, nil, top]
active₁ > [cons₂, sel₂] > indx₂ > [ok₁, dbl₁, nil, top]
proper₁ > 0 > [ok₁, dbl₁, nil, top]
proper₁ > [cons₂, sel₂] > indx₂ > [ok₁, dbl₁, nil, top]

Status:

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

The following usable rules [FROCOS05] were oriented:

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

(29) 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)

The TRS R consists of the following rules:

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

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(X1, mark(X2)) → SEL(X1, X2)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[active₁, indx₂] > [cons₂, sel₂] > [mark₁, 0, nil, ok, top]

Status:

mark₁: [1]
active₁: [1]
0: []
nil: []
cons₂: [2,1]
sel₂: [1,2]
indx₂: [1,2]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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

(31) Obligation:

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

SEL(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(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

proper₁ > [active₁, indx₂] > [cons₂, sel₂] > [mark₁, 0, ok₁, top]
proper₁ > nil > [mark₁, 0, ok₁, top]

Status:

mark₁: [1]
active₁: [1]
0: []
nil: []
cons₂: [2,1]
sel₂: [2,1]
indx₂: [2,1]
proper₁: [1]
ok₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(33) Obligation:

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

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

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

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBLS(ok(X)) → DBLS(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

DBLS₁ > [ok₁, dbl₁, nil, top]
[active₁, indx₁, proper₁] > 0 > [ok₁, dbl₁, nil, top]
[active₁, indx₁, proper₁] > [cons₂, sel₁] > [ok₁, dbl₁, nil, top]

Status:

DBLS₁: [1]
ok₁: [1]
active₁: [1]
dbl₁: [1]
0: []
nil: []
cons₂: [1,2]
sel₁: [1]
indx₁: [1]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(38) Obligation:

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

DBLS(mark(X)) → DBLS(X)

The TRS R consists of the following rules:

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

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

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBLS(mark(X)) → DBLS(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[active₁, indx₂, proper₁] > [cons₂, sel₂] > [mark₁, 0, from₁, ok₁, top]
nil > [mark₁, 0, from₁, ok₁, top]

Status:

mark₁: [1]
active₁: [1]
0: []
nil: []
cons₂: [2,1]
sel₂: [2,1]
indx₂: [2,1]
from₁: [1]
proper₁: [1]
ok₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(40) Obligation:

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

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

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

(41) PisEmptyProof (EQUIVALENT transformation)

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

(42) TRUE

(43) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(ok(X)) → DBL(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

DBL₁ > [ok₁, dbl₁, nil, top]
[active₁, indx₁, proper₁] > 0 > [ok₁, dbl₁, nil, top]
[active₁, indx₁, proper₁] > [cons₂, sel₁] > [ok₁, dbl₁, nil, top]

Status:

DBL₁: [1]
ok₁: [1]
active₁: [1]
dbl₁: [1]
0: []
nil: []
cons₂: [1,2]
sel₁: [1]
indx₁: [1]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(45) Obligation:

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

DBL(mark(X)) → DBL(X)

The TRS R consists of the following rules:

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

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

(46) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(mark(X)) → DBL(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[active₁, indx₂, proper₁] > [cons₂, sel₂] > [mark₁, 0, from₁, ok₁, top]
nil > [mark₁, 0, from₁, ok₁, top]

Status:

mark₁: [1]
active₁: [1]
0: []
nil: []
cons₂: [2,1]
sel₂: [2,1]
indx₂: [2,1]
from₁: [1]
proper₁: [1]
ok₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(47) Obligation:

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

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

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

(48) PisEmptyProof (EQUIVALENT transformation)

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

(49) TRUE

(50) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(s(X)) → PROPER(X)
PROPER(dbl(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(indx(X1, X2)) → PROPER(X1)
PROPER(indx(X1, X2)) → PROPER(X2)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[cons₂, sel₂] > indx₂ > [s₁, dbl₁, mark, ok, top]
0 > [s₁, dbl₁, mark, ok, top]
nil > [s₁, dbl₁, mark, ok, top]

Status:

s₁: [1]
dbl₁: [1]
cons₂: [1,2]
sel₂: [2,1]
indx₂: [1,2]
0: []
mark: []
nil: []
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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

(52) Obligation:

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

PROPER(dbls(X)) → PROPER(X)
PROPER(from(X)) → PROPER(X)

The TRS R consists of the following rules:

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

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

(53) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(from(X)) → PROPER(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[from₁, active₁] > indx₂ > [dbl₁, mark, s₁, nil, cons₁, sel₂, ok, top]
0 > [dbl₁, mark, s₁, nil, cons₁, sel₂, ok, top]

Status:

from₁: [1]
active₁: [1]
dbl₁: [1]
0: []
mark: []
s₁: [1]
nil: []
cons₁: [1]
sel₂: [2,1]
indx₂: [2,1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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

(54) Obligation:

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

PROPER(dbls(X)) → PROPER(X)

The TRS R consists of the following rules:

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

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

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(dbls(X)) → PROPER(X)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

dbls₁ > PROPER₁ > [dbl₁, mark, s₁, cons₂, sel₂, top]
0 > [dbl₁, mark, s₁, cons₂, sel₂, top]
nil > [dbl₁, mark, s₁, cons₂, sel₂, top]
indx₂ > [dbl₁, mark, s₁, cons₂, sel₂, top]
from₁ > [dbl₁, mark, s₁, cons₂, sel₂, top]

Status:

PROPER₁: [1]
dbls₁: [1]
dbl₁: [1]
0: []
mark: []
s₁: [1]
nil: []
cons₂: [2,1]
sel₂: [1,2]
indx₂: [2,1]
from₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(56) Obligation:

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

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

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

(57) PisEmptyProof (EQUIVALENT transformation)

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

(58) TRUE

(59) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(indx(X1, X2)) → ACTIVE(X1)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[ACTIVE₁, sel₂] > [mark, s₁, cons₂, ok, top]
indx₁ > nil > [mark, s₁, cons₂, ok, top]
0 > [mark, s₁, cons₂, ok, top]
from₁ > [mark, s₁, cons₂, ok, top]

Status:

ACTIVE₁: [1]
sel₂: [1,2]
indx₁: [1]
0: []
mark: []
s₁: [1]
nil: []
cons₂: [2,1]
from₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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

(61) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

dbls₁ > [dbl₁, mark, s₁, cons₂, sel₂, ok, top]
0 > [dbl₁, mark, s₁, cons₂, sel₂, ok, top]
nil > [dbl₁, mark, s₁, cons₂, sel₂, ok, top]
indx₂ > [dbl₁, mark, s₁, cons₂, sel₂, ok, top]
from₁ > [dbl₁, mark, s₁, cons₂, sel₂, ok, top]

Status:

dbls₁: [1]
dbl₁: [1]
0: []
mark: []
s₁: [1]
nil: []
cons₂: [2,1]
sel₂: [2,1]
indx₂: [2,1]
from₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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

(63) Obligation:

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

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

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

(64) PisEmptyProof (EQUIVALENT transformation)

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

(65) TRUE

(66) Obligation:

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

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

The TRS R consists of the following rules:

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

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