(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(nil, YS)) → MARK(YS)
ACTIVE(app(cons(X, XS), YS)) → MARK(cons(X, app(XS, YS)))
ACTIVE(app(cons(X, XS), YS)) → CONS(X, app(XS, YS))
ACTIVE(app(cons(X, XS), YS)) → APP(XS, YS)
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)
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(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)) → MARK(cons(nil, zWadr(L, prefix(L))))
ACTIVE(prefix(L)) → CONS(nil, zWadr(L, prefix(L)))
ACTIVE(prefix(L)) → ZWADR(L, prefix(L))
MARK(app(X1, X2)) → ACTIVE(app(mark(X1), mark(X2)))
MARK(app(X1, X2)) → APP(mark(X1), mark(X2))
MARK(app(X1, X2)) → MARK(X1)
MARK(app(X1, X2)) → MARK(X2)
MARK(nil) → ACTIVE(nil)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(zWadr(X1, X2)) → ACTIVE(zWadr(mark(X1), mark(X2)))
MARK(zWadr(X1, X2)) → ZWADR(mark(X1), mark(X2))
MARK(zWadr(X1, X2)) → MARK(X1)
MARK(zWadr(X1, X2)) → MARK(X2)
MARK(prefix(X)) → ACTIVE(prefix(mark(X)))
MARK(prefix(X)) → PREFIX(mark(X))
MARK(prefix(X)) → MARK(X)
APP(mark(X1), X2) → APP(X1, X2)
APP(X1, mark(X2)) → APP(X1, X2)
APP(active(X1), X2) → APP(X1, X2)
APP(X1, active(X2)) → APP(X1, X2)
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)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
ZWADR(mark(X1), X2) → ZWADR(X1, X2)
ZWADR(X1, mark(X2)) → ZWADR(X1, X2)
ZWADR(active(X1), X2) → ZWADR(X1, X2)
ZWADR(X1, active(X2)) → ZWADR(X1, X2)
PREFIX(mark(X)) → PREFIX(X)
PREFIX(active(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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 7 SCCs with 20 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

PREFIX(active(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(active(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
active(x1)  =  active(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(active(X1), X2) → ZWADR(X1, X2)
ZWADR(X1, active(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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, active(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)  =  x2
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(14) 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(active(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(mark(X1), X2) → ZWADR(X1, X2)
ZWADR(active(X1), 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)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

ZWADR(X1, mark(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(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)  =  x2
mark(x1)  =  mark(x1)

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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(active(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(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

Lexicographic Path Order [LPO].
Precedence:
trivial


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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(active(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(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

Lexicographic Path Order [LPO].
Precedence:
trivial


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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(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(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(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)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(37) 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)

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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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)
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)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(39) Obligation:

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

CONS(X1, mark(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(40) 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)
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)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(41) 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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) 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(active(X1), X2) → APP(X1, X2)
APP(X1, active(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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(X1, active(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)  =  x2
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(46) 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(active(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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)
APP(active(X1), 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)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

APP(X1, mark(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(50) Obligation:

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

active(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) Obligation:

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

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


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

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


The following usable rules [FROCOS05] were oriented:

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

(55) Obligation:

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

MARK(app(X1, X2)) → ACTIVE(app(mark(X1), mark(X2)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(zWadr(X1, X2)) → ACTIVE(zWadr(mark(X1), mark(X2)))
ACTIVE(prefix(L)) → MARK(cons(nil, zWadr(L, prefix(L))))
MARK(prefix(X)) → ACTIVE(prefix(mark(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(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.


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

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


The following usable rules [FROCOS05] were oriented:

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

(57) Obligation:

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

MARK(app(X1, X2)) → ACTIVE(app(mark(X1), mark(X2)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(zWadr(X1, X2)) → ACTIVE(zWadr(mark(X1), mark(X2)))
MARK(prefix(X)) → ACTIVE(prefix(mark(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))))
mark(app(X1, X2)) → active(app(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(zWadr(X1, X2)) → active(zWadr(mark(X1), mark(X2)))
mark(prefix(X)) → active(prefix(mark(X)))
app(mark(X1), X2) → app(X1, X2)
app(X1, mark(X2)) → app(X1, X2)
app(active(X1), X2) → app(X1, X2)
app(X1, active(X2)) → app(X1, X2)
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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
zWadr(mark(X1), X2) → zWadr(X1, X2)
zWadr(X1, mark(X2)) → zWadr(X1, X2)
zWadr(active(X1), X2) → zWadr(X1, X2)
zWadr(X1, active(X2)) → zWadr(X1, X2)
prefix(mark(X)) → prefix(X)
prefix(active(X)) → prefix(X)

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

(58) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 6 less nodes.

(59) TRUE