Termination w.r.t. Q of the given QTRS could be proven:

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 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 PisEmptyProof (⇔)
    ↳27 TRUE
  ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 QDPOrderProof (⇔)
    ↳32 QDP
    ↳33 PisEmptyProof (⇔)
    ↳34 TRUE
  ↳35 QDP
    ↳36 QDPOrderProof (⇔)
    ↳37 QDP
    ↳38 QDPOrderProof (⇔)
    ↳39 QDP
    ↳40 PisEmptyProof (⇔)
    ↳41 TRUE
  ↳42 QDP
    ↳43 QDPOrderProof (⇔)
    ↳44 QDP
    ↳45 QDPOrderProof (⇔)
    ↳46 QDP
    ↳47 QDPOrderProof (⇔)
    ↳48 QDP
    ↳49 PisEmptyProof (⇔)
    ↳50 TRUE
  ↳51 QDP
    ↳52 QDPOrderProof (⇔)
    ↳53 QDP
    ↳54 QDPOrderProof (⇔)
    ↳55 QDP
    ↳56 QDPOrderProof (⇔)
    ↳57 QDP
    ↳58 PisEmptyProof (⇔)
    ↳59 TRUE
  ↳60 QDP
    ↳61 QDPOrderProof (⇔)
    ↳62 QDP
    ↳63 QDPOrderProof (⇔)
    ↳64 QDP
    ↳65 QDPOrderProof (⇔)
    ↳66 QDP
    ↳67 QDPOrderProof (⇔)
    ↳68 QDP
    ↳69 PisEmptyProof (⇔)
    ↳70 TRUE
  ↳71 QDP
    ↳72 QDPOrderProof (⇔)
    ↳73 QDP
    ↳74 QDPOrderProof (⇔)
    ↳75 QDP
    ↳76 PisEmptyProof (⇔)
    ↳77 TRUE

(0) Obligation:

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

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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(app(cons(X, XS), YS)) → CONS(X, app(XS, YS))
ACTIVE(app(cons(X, XS), YS)) → APP(XS, YS)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → CONS(app(Y, cons(X, nil)), zWadr(XS, YS))
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → APP(Y, cons(X, nil))
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → CONS(X, nil)
ACTIVE(zWadr(cons(X, XS), cons(Y, YS))) → ZWADR(XS, YS)
ACTIVE(prefix(L)) → CONS(nil, zWadr(L, prefix(L)))
ACTIVE(prefix(L)) → ZWADR(L, prefix(L))
ACTIVE(app(X1, X2)) → APP(active(X1), X2)
ACTIVE(app(X1, X2)) → ACTIVE(X1)
ACTIVE(app(X1, X2)) → APP(X1, active(X2))
ACTIVE(app(X1, X2)) → ACTIVE(X2)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(zWadr(X1, X2)) → ZWADR(active(X1), X2)
ACTIVE(zWadr(X1, X2)) → ACTIVE(X1)
ACTIVE(zWadr(X1, X2)) → ZWADR(X1, active(X2))
ACTIVE(zWadr(X1, X2)) → ACTIVE(X2)
ACTIVE(prefix(X)) → PREFIX(active(X))
ACTIVE(prefix(X)) → ACTIVE(X)
APP(mark(X1), X2) → APP(X1, X2)
APP(X1, mark(X2)) → APP(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
S(mark(X)) → S(X)
ZWADR(mark(X1), X2) → ZWADR(X1, X2)
ZWADR(X1, mark(X2)) → ZWADR(X1, X2)
PREFIX(mark(X)) → PREFIX(X)
PROPER(app(X1, X2)) → APP(proper(X1), proper(X2))
PROPER(app(X1, X2)) → PROPER(X1)
PROPER(app(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → FROM(proper(X))
PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(zWadr(X1, X2)) → ZWADR(proper(X1), proper(X2))
PROPER(zWadr(X1, X2)) → PROPER(X1)
PROPER(zWadr(X1, X2)) → PROPER(X2)
PROPER(prefix(X)) → PREFIX(proper(X))
PROPER(prefix(X)) → PROPER(X)
APP(ok(X1), ok(X2)) → APP(X1, X2)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
FROM(ok(X)) → FROM(X)
S(ok(X)) → S(X)
ZWADR(ok(X1), ok(X2)) → ZWADR(X1, X2)
PREFIX(ok(X)) → PREFIX(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(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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 9 SCCs with 27 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

PREFIX(ok(X)) → PREFIX(X)
PREFIX(mark(X)) → PREFIX(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.


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

Lexicographic Path Order [LPO].
Precedence:
[PREFIX1, ok1]


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

PREFIX(mark(X)) → PREFIX(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


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(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(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:

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

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
mark1 > ZWADR1


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
ok1 > [ZWADR1, mark]


The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

ZWADR(mark(X1), X2) → ZWADR(X1, X2)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ZWADR(mark(X1), X2) → ZWADR(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) Obligation:

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
[S1, ok1]


The following usable rules [FROCOS05] were oriented: none

(23) 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(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(24) 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: Combined order from the following AFS and order.
S(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE

(28) Obligation:

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

FROM(ok(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
[FROM1, ok1]


The following usable rules [FROCOS05] were oriented: none

(30) 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(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(31) 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: Combined order from the following AFS and order.
FROM(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

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

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(33) PisEmptyProof (EQUIVALENT transformation)

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

(34) TRUE

(35) Obligation:

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

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

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(36) 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)
mark(x1)  =  mark

Lexicographic Path Order [LPO].
Precedence:
ok1 > CONS1
mark > CONS1


The following usable rules [FROCOS05] were oriented: none

(37) Obligation:

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

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(38) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(40) PisEmptyProof (EQUIVALENT transformation)

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

(41) TRUE

(42) Obligation:

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

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

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
mark1 > APP1


The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

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

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

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
ok1 > [APP1, mark]


The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

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

APP(mark(X1), X2) → APP(X1, X2)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(mark(X1), X2) → APP(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(49) PisEmptyProof (EQUIVALENT transformation)

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

(50) TRUE

(51) Obligation:

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

PROPER(app(X1, X2)) → PROPER(X2)
PROPER(app(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(zWadr(X1, X2)) → PROPER(X1)
PROPER(zWadr(X1, X2)) → PROPER(X2)
PROPER(prefix(X)) → PROPER(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(app(X1, X2)) → PROPER(X2)
PROPER(app(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(zWadr(X1, X2)) → PROPER(X1)
PROPER(zWadr(X1, X2)) → PROPER(X2)
PROPER(prefix(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)
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
from(x1)  =  x1
s(x1)  =  x1
zWadr(x1, x2)  =  zWadr(x1, x2)
prefix(x1)  =  prefix(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(53) Obligation:

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

PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
[PROPER1, from1]


The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

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

PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

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

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(58) PisEmptyProof (EQUIVALENT transformation)

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

(59) TRUE

(60) Obligation:

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

ACTIVE(app(X1, X2)) → ACTIVE(X2)
ACTIVE(app(X1, X2)) → ACTIVE(X1)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(zWadr(X1, X2)) → ACTIVE(X1)
ACTIVE(zWadr(X1, X2)) → ACTIVE(X2)
ACTIVE(prefix(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(app(X1, X2)) → ACTIVE(X2)
ACTIVE(app(X1, X2)) → ACTIVE(X1)
ACTIVE(zWadr(X1, X2)) → ACTIVE(X1)
ACTIVE(zWadr(X1, X2)) → ACTIVE(X2)
ACTIVE(prefix(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  x1
from(x1)  =  x1
s(x1)  =  x1
zWadr(x1, x2)  =  zWadr(x1, x2)
prefix(x1)  =  prefix(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(62) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(s(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
cons(x1, x2)  =  x1
from(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(64) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(65) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
from1 > ACTIVE1


The following usable rules [FROCOS05] were oriented: none

(66) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(67) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(cons(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)
cons(x1, x2)  =  cons(x1)

Lexicographic Path Order [LPO].
Precedence:
[ACTIVE1, cons1]


The following usable rules [FROCOS05] were oriented: none

(68) Obligation:

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

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(69) PisEmptyProof (EQUIVALENT transformation)

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

(70) TRUE

(71) 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(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(72) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(mark(X)) → TOP(proper(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
ok(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)
proper(x1)  =  x1
zWadr(x1, x2)  =  zWadr(x1, x2)
cons(x1, x2)  =  cons(x1)
app(x1, x2)  =  app(x1, x2)
nil  =  nil
prefix(x1)  =  prefix(x1)
from(x1)  =  from(x1)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
TOP1 > mark1
prefix1 > zWadr2 > [cons1, app2] > mark1
prefix1 > zWadr2 > nil > mark1
[from1, s1] > [cons1, app2] > mark1


The following usable rules [FROCOS05] were oriented:

active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
active(app(nil, YS)) → mark(YS)
active(from(X)) → mark(cons(X, from(s(X))))
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(nil, YS)) → mark(nil)
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(app(X1, X2)) → app(proper(X1), proper(X2))
prefix(mark(X)) → mark(prefix(X))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
s(ok(X)) → ok(s(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
proper(prefix(X)) → prefix(proper(X))

(73) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(74) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(ok(X)) → TOP(active(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1)  =  TOP(x1)
ok(x1)  =  ok(x1)
active(x1)  =  x1
zWadr(x1, x2)  =  x2
cons(x1, x2)  =  x2
mark(x1)  =  x1
app(x1, x2)  =  app(x2)
nil  =  nil
prefix(x1)  =  x1
from(x1)  =  x1
s(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
[TOP1, ok1, app1] > nil


The following usable rules [FROCOS05] were oriented:

active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
active(app(nil, YS)) → mark(YS)
active(from(X)) → mark(cons(X, from(s(X))))
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(nil, YS)) → mark(nil)
prefix(mark(X)) → mark(prefix(X))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
s(ok(X)) → ok(s(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
app(ok(X1), ok(X2)) → ok(app(X1, X2))

(75) Obligation:

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

active(app(nil, YS)) → mark(YS)
active(app(cons(X, XS), YS)) → mark(cons(X, app(XS, YS)))
active(from(X)) → mark(cons(X, from(s(X))))
active(zWadr(nil, YS)) → mark(nil)
active(zWadr(XS, nil)) → mark(nil)
active(zWadr(cons(X, XS), cons(Y, YS))) → mark(cons(app(Y, cons(X, nil)), zWadr(XS, YS)))
active(prefix(L)) → mark(cons(nil, zWadr(L, prefix(L))))
active(app(X1, X2)) → app(active(X1), X2)
active(app(X1, X2)) → app(X1, active(X2))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(zWadr(X1, X2)) → zWadr(active(X1), X2)
active(zWadr(X1, X2)) → zWadr(X1, active(X2))
active(prefix(X)) → prefix(active(X))
app(mark(X1), X2) → mark(app(X1, X2))
app(X1, mark(X2)) → mark(app(X1, X2))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
zWadr(mark(X1), X2) → mark(zWadr(X1, X2))
zWadr(X1, mark(X2)) → mark(zWadr(X1, X2))
prefix(mark(X)) → mark(prefix(X))
proper(app(X1, X2)) → app(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(zWadr(X1, X2)) → zWadr(proper(X1), proper(X2))
proper(prefix(X)) → prefix(proper(X))
app(ok(X1), ok(X2)) → ok(app(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
zWadr(ok(X1), ok(X2)) → ok(zWadr(X1, X2))
prefix(ok(X)) → ok(prefix(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.

(76) PisEmptyProof (EQUIVALENT transformation)

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

(77) TRUE