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 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 TRUE
  ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 PisEmptyProof (⇔)
    ↳25 TRUE
  ↳26 QDP
    ↳27 QDPOrderProof (⇔)
    ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 PisEmptyProof (⇔)
    ↳32 TRUE
  ↳33 QDP
    ↳34 QDPOrderProof (⇔)
    ↳35 QDP
    ↳36 QDPOrderProof (⇔)
    ↳37 QDP
    ↳38 PisEmptyProof (⇔)
    ↳39 TRUE
  ↳40 QDP
    ↳41 QDPOrderProof (⇔)
    ↳42 QDP
    ↳43 QDPOrderProof (⇔)
    ↳44 QDP
    ↳45 PisEmptyProof (⇔)
    ↳46 TRUE
  ↳47 QDP
    ↳48 QDPOrderProof (⇔)
    ↳49 QDP
    ↳50 QDPOrderProof (⇔)
    ↳51 QDP
    ↳52 PisEmptyProof (⇔)
    ↳53 TRUE
  ↳54 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(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(0)) → MARK(0)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
ACTIVE(dbl(s(X))) → S(s(dbl(X)))
ACTIVE(dbl(s(X))) → S(dbl(X))
ACTIVE(dbl(s(X))) → DBL(X)
ACTIVE(dbls(nil)) → MARK(nil)
ACTIVE(dbls(cons(X, Y))) → MARK(cons(dbl(X), dbls(Y)))
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(0, cons(X, Y))) → MARK(X)
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(indx(nil, X)) → MARK(nil)
ACTIVE(indx(cons(X, Y), Z)) → MARK(cons(sel(X, Z), indx(Y, 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)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
MARK(dbl(X)) → DBL(mark(X))
MARK(dbl(X)) → MARK(X)
MARK(0) → ACTIVE(0)
MARK(s(X)) → ACTIVE(s(X))
MARK(dbls(X)) → ACTIVE(dbls(mark(X)))
MARK(dbls(X)) → DBLS(mark(X))
MARK(dbls(X)) → MARK(X)
MARK(nil) → ACTIVE(nil)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(sel(X1, X2)) → SEL(mark(X1), mark(X2))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
MARK(indx(X1, X2)) → ACTIVE(indx(mark(X1), X2))
MARK(indx(X1, X2)) → INDX(mark(X1), X2)
MARK(indx(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(X))
DBL(mark(X)) → DBL(X)
DBL(active(X)) → DBL(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
DBLS(mark(X)) → DBLS(X)
DBLS(active(X)) → DBLS(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)
INDX(mark(X1), X2) → INDX(X1, X2)
INDX(X1, mark(X2)) → INDX(X1, X2)
INDX(active(X1), X2) → INDX(X1, X2)
INDX(X1, active(X2)) → INDX(X1, X2)
FROM(mark(X)) → FROM(X)
FROM(active(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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(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 8 SCCs with 22 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

FROM(active(X)) → FROM(X)
FROM(mark(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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(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(active(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
active(x1)  =  active(x1)
mark(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

FROM(mark(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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FROM(mark(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:
mark1 > FROM1

The following usable rules [FROCOS05] were oriented: none

(9) 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

INDX(X1, mark(X2)) → INDX(X1, X2)
INDX(mark(X1), X2) → INDX(X1, X2)
INDX(active(X1), X2) → INDX(X1, X2)
INDX(X1, active(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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


INDX(X1, mark(X2)) → INDX(X1, X2)
INDX(X1, active(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)  =  x2
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)

Recursive Path Order [RPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

INDX(mark(X1), X2) → INDX(X1, X2)
INDX(active(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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(15) 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)
INDX(active(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)

Recursive Path Order [RPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(16) 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

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

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

(20) 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(X1, active(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)

Recursive Path Order [RPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

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

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

(22) 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)
SEL(active(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)

Recursive Path Order [RPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(23) 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(X1, active(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)  =  x2
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)

Recursive Path Order [RPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), 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)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)

Recursive Path Order [RPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(30) 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

DBLS(active(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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DBLS(active(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
active(x1)  =  active(x1)
mark(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(35) 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(36) 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: Recursive Path Order [RPO].
Precedence:
mark1 > DBLS1

The following usable rules [FROCOS05] were oriented: none

(37) 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) Obligation:

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

S(active(X)) → S(X)
S(mark(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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(active(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
active(x1)  =  active(x1)
mark(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(42) Obligation:

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

S(mark(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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:
mark1 > S1

The following usable rules [FROCOS05] were oriented: none

(44) 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) Obligation:

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

DBL(active(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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(48) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DBL(active(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
active(x1)  =  active(x1)
mark(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(49) 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(50) 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: Recursive Path Order [RPO].
Precedence:
mark1 > DBL1

The following usable rules [FROCOS05] were oriented: none

(51) 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))))
mark(dbl(X)) → active(dbl(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(dbls(X)) → active(dbls(mark(X)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(indx(X1, X2)) → active(indx(mark(X1), X2))
mark(from(X)) → active(from(X))
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
dbls(mark(X)) → dbls(X)
dbls(active(X)) → dbls(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
indx(mark(X1), X2) → indx(X1, X2)
indx(X1, mark(X2)) → indx(X1, X2)
indx(active(X1), X2) → indx(X1, X2)
indx(X1, active(X2)) → indx(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(52) PisEmptyProof (EQUIVALENT transformation)

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

(53) TRUE

(54) Obligation:

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

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

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