Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.

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

ACTIVE1(zWadr2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(from1(X)) -> FROM1(s1(X))
PROPER1(from1(X)) -> FROM1(proper1(X))
ACTIVE1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> CONS2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
ACTIVE1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> APP2(Y, cons2(X, nil))
ACTIVE1(from1(X)) -> FROM1(active1(X))
S1(mark1(X)) -> S1(X)
PROPER1(prefix1(X)) -> PREFIX1(proper1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(app2(X1, X2)) -> APP2(proper1(X1), proper1(X2))
ZWADR2(X1, mark1(X2)) -> ZWADR2(X1, X2)
ACTIVE1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> ZWADR2(XS, YS)
PROPER1(zWadr2(X1, X2)) -> ZWADR2(proper1(X1), proper1(X2))
ACTIVE1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> CONS2(X, nil)
ZWADR2(mark1(X1), X2) -> ZWADR2(X1, X2)
PROPER1(zWadr2(X1, X2)) -> PROPER1(X2)
ACTIVE1(zWadr2(X1, X2)) -> ZWADR2(active1(X1), X2)
ACTIVE1(app2(X1, X2)) -> APP2(X1, active1(X2))
ACTIVE1(prefix1(L)) -> ZWADR2(L, prefix1(L))
FROM1(mark1(X)) -> FROM1(X)
PREFIX1(mark1(X)) -> PREFIX1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(app2(X1, X2)) -> APP2(active1(X1), X2)
APP2(ok1(X1), ok1(X2)) -> APP2(X1, X2)
ACTIVE1(zWadr2(X1, X2)) -> ACTIVE1(X2)
PROPER1(app2(X1, X2)) -> PROPER1(X2)
TOP1(ok1(X)) -> ACTIVE1(X)
APP2(mark1(X1), X2) -> APP2(X1, X2)
ACTIVE1(s1(X)) -> S1(active1(X))
ACTIVE1(app2(cons2(X, XS), YS)) -> CONS2(X, app2(XS, YS))
PROPER1(app2(X1, X2)) -> PROPER1(X1)
ACTIVE1(app2(X1, X2)) -> ACTIVE1(X2)
PROPER1(s1(X)) -> PROPER1(X)
TOP1(mark1(X)) -> PROPER1(X)
APP2(X1, mark1(X2)) -> APP2(X1, X2)
ACTIVE1(prefix1(L)) -> CONS2(nil, zWadr2(L, prefix1(L)))
PROPER1(zWadr2(X1, X2)) -> PROPER1(X1)
ZWADR2(ok1(X1), ok1(X2)) -> ZWADR2(X1, X2)
ACTIVE1(zWadr2(X1, X2)) -> ZWADR2(X1, active1(X2))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(from1(X)) -> CONS2(X, from1(s1(X)))
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(prefix1(X)) -> PREFIX1(active1(X))
PROPER1(prefix1(X)) -> PROPER1(X)
ACTIVE1(from1(X)) -> S1(X)
ACTIVE1(prefix1(X)) -> ACTIVE1(X)
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PREFIX1(ok1(X)) -> PREFIX1(X)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
FROM1(ok1(X)) -> FROM1(X)
ACTIVE1(app2(cons2(X, XS), YS)) -> APP2(XS, YS)
ACTIVE1(app2(X1, X2)) -> ACTIVE1(X1)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

ACTIVE1(zWadr2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(from1(X)) -> FROM1(s1(X))
PROPER1(from1(X)) -> FROM1(proper1(X))
ACTIVE1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> CONS2(app2(Y, cons2(X, nil)), zWadr2(XS, YS))
ACTIVE1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> APP2(Y, cons2(X, nil))
ACTIVE1(from1(X)) -> FROM1(active1(X))
S1(mark1(X)) -> S1(X)
PROPER1(prefix1(X)) -> PREFIX1(proper1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(app2(X1, X2)) -> APP2(proper1(X1), proper1(X2))
ZWADR2(X1, mark1(X2)) -> ZWADR2(X1, X2)
ACTIVE1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> ZWADR2(XS, YS)
PROPER1(zWadr2(X1, X2)) -> ZWADR2(proper1(X1), proper1(X2))
ACTIVE1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> CONS2(X, nil)
ZWADR2(mark1(X1), X2) -> ZWADR2(X1, X2)
PROPER1(zWadr2(X1, X2)) -> PROPER1(X2)
ACTIVE1(zWadr2(X1, X2)) -> ZWADR2(active1(X1), X2)
ACTIVE1(app2(X1, X2)) -> APP2(X1, active1(X2))
ACTIVE1(prefix1(L)) -> ZWADR2(L, prefix1(L))
FROM1(mark1(X)) -> FROM1(X)
PREFIX1(mark1(X)) -> PREFIX1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(app2(X1, X2)) -> APP2(active1(X1), X2)
APP2(ok1(X1), ok1(X2)) -> APP2(X1, X2)
ACTIVE1(zWadr2(X1, X2)) -> ACTIVE1(X2)
PROPER1(app2(X1, X2)) -> PROPER1(X2)
TOP1(ok1(X)) -> ACTIVE1(X)
APP2(mark1(X1), X2) -> APP2(X1, X2)
ACTIVE1(s1(X)) -> S1(active1(X))
ACTIVE1(app2(cons2(X, XS), YS)) -> CONS2(X, app2(XS, YS))
PROPER1(app2(X1, X2)) -> PROPER1(X1)
ACTIVE1(app2(X1, X2)) -> ACTIVE1(X2)
PROPER1(s1(X)) -> PROPER1(X)
TOP1(mark1(X)) -> PROPER1(X)
APP2(X1, mark1(X2)) -> APP2(X1, X2)
ACTIVE1(prefix1(L)) -> CONS2(nil, zWadr2(L, prefix1(L)))
PROPER1(zWadr2(X1, X2)) -> PROPER1(X1)
ZWADR2(ok1(X1), ok1(X2)) -> ZWADR2(X1, X2)
ACTIVE1(zWadr2(X1, X2)) -> ZWADR2(X1, active1(X2))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(from1(X)) -> CONS2(X, from1(s1(X)))
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(prefix1(X)) -> PREFIX1(active1(X))
PROPER1(prefix1(X)) -> PROPER1(X)
ACTIVE1(from1(X)) -> S1(X)
ACTIVE1(prefix1(X)) -> ACTIVE1(X)
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PREFIX1(ok1(X)) -> PREFIX1(X)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
FROM1(ok1(X)) -> FROM1(X)
ACTIVE1(app2(cons2(X, XS), YS)) -> APP2(XS, YS)
ACTIVE1(app2(X1, X2)) -> ACTIVE1(X1)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 9 SCCs with 27 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

PREFIX1(mark1(X)) -> PREFIX1(X)
PREFIX1(ok1(X)) -> PREFIX1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PREFIX1(ok1(X)) -> PREFIX1(X)
Used argument filtering: PREFIX1(x1)  =  x1
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

PREFIX1(mark1(X)) -> PREFIX1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PREFIX1(mark1(X)) -> PREFIX1(X)
Used argument filtering: PREFIX1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ZWADR2(mark1(X1), X2) -> ZWADR2(X1, X2)
ZWADR2(ok1(X1), ok1(X2)) -> ZWADR2(X1, X2)
ZWADR2(X1, mark1(X2)) -> ZWADR2(X1, X2)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ZWADR2(X1, mark1(X2)) -> ZWADR2(X1, X2)
Used argument filtering: ZWADR2(x1, x2)  =  x2
ok1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ZWADR2(mark1(X1), X2) -> ZWADR2(X1, X2)
ZWADR2(ok1(X1), ok1(X2)) -> ZWADR2(X1, X2)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ZWADR2(ok1(X1), ok1(X2)) -> ZWADR2(X1, X2)
Used argument filtering: ZWADR2(x1, x2)  =  x2
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ZWADR2(mark1(X1), X2) -> ZWADR2(X1, X2)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ZWADR2(mark1(X1), X2) -> ZWADR2(X1, X2)
Used argument filtering: ZWADR2(x1, x2)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

S1(ok1(X)) -> S1(X)
S1(mark1(X)) -> S1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S1(mark1(X)) -> S1(X)
Used argument filtering: S1(x1)  =  x1
ok1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

S1(ok1(X)) -> S1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S1(ok1(X)) -> S1(X)
Used argument filtering: S1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

FROM1(mark1(X)) -> FROM1(X)
FROM1(ok1(X)) -> FROM1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FROM1(ok1(X)) -> FROM1(X)
Used argument filtering: FROM1(x1)  =  x1
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

FROM1(mark1(X)) -> FROM1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FROM1(mark1(X)) -> FROM1(X)
Used argument filtering: FROM1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2)  =  x2
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

CONS2(mark1(X1), X2) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP2(mark1(X1), X2) -> APP2(X1, X2)
APP2(X1, mark1(X2)) -> APP2(X1, X2)
APP2(ok1(X1), ok1(X2)) -> APP2(X1, X2)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(ok1(X1), ok1(X2)) -> APP2(X1, X2)
Used argument filtering: APP2(x1, x2)  =  x2
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP2(mark1(X1), X2) -> APP2(X1, X2)
APP2(X1, mark1(X2)) -> APP2(X1, X2)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(X1, mark1(X2)) -> APP2(X1, X2)
Used argument filtering: APP2(x1, x2)  =  x2
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

APP2(mark1(X1), X2) -> APP2(X1, X2)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(mark1(X1), X2) -> APP2(X1, X2)
Used argument filtering: APP2(x1, x2)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(prefix1(X)) -> PROPER1(X)
PROPER1(zWadr2(X1, X2)) -> PROPER1(X2)
PROPER1(app2(X1, X2)) -> PROPER1(X1)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(zWadr2(X1, X2)) -> PROPER1(X1)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(app2(X1, X2)) -> PROPER1(X2)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(zWadr2(X1, X2)) -> PROPER1(X2)
PROPER1(app2(X1, X2)) -> PROPER1(X1)
PROPER1(zWadr2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(app2(X1, X2)) -> PROPER1(X2)
Used argument filtering: PROPER1(x1)  =  x1
prefix1(x1)  =  x1
zWadr2(x1, x2)  =  zWadr2(x1, x2)
app2(x1, x2)  =  app2(x1, x2)
s1(x1)  =  x1
from1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(prefix1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(from1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
prefix1(x1)  =  x1
s1(x1)  =  x1
from1(x1)  =  from1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(prefix1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(s1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
prefix1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(prefix1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(prefix1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
prefix1(x1)  =  prefix1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
QDP
                            ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(zWadr2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(prefix1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(app2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
ACTIVE1(zWadr2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(app2(X1, X2)) -> ACTIVE1(X1)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(zWadr2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(app2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(zWadr2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(app2(X1, X2)) -> ACTIVE1(X1)
Used argument filtering: ACTIVE1(x1)  =  x1
zWadr2(x1, x2)  =  zWadr2(x1, x2)
prefix1(x1)  =  x1
cons2(x1, x2)  =  x1
app2(x1, x2)  =  app2(x1, x2)
from1(x1)  =  x1
s1(x1)  =  x1
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(prefix1(X)) -> ACTIVE1(X)
ACTIVE1(from1(X)) -> ACTIVE1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(s1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
cons2(x1, x2)  =  x1
prefix1(x1)  =  x1
from1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(prefix1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(from1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(from1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
prefix1(x1)  =  x1
cons2(x1, x2)  =  x1
from1(x1)  =  from1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(prefix1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(prefix1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
cons2(x1, x2)  =  x1
prefix1(x1)  =  prefix1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
QDP
                            ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
Used argument filtering: ACTIVE1(x1)  =  x1
cons2(x1, x2)  =  cons1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
                          ↳ QDP
                            ↳ QDPAfsSolverProof
QDP
                                ↳ PisEmptyProof
          ↳ QDP

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

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP

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

TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))

The TRS R consists of the following rules:

active1(app2(nil, YS)) -> mark1(YS)
active1(app2(cons2(X, XS), YS)) -> mark1(cons2(X, app2(XS, YS)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(zWadr2(nil, YS)) -> mark1(nil)
active1(zWadr2(XS, nil)) -> mark1(nil)
active1(zWadr2(cons2(X, XS), cons2(Y, YS))) -> mark1(cons2(app2(Y, cons2(X, nil)), zWadr2(XS, YS)))
active1(prefix1(L)) -> mark1(cons2(nil, zWadr2(L, prefix1(L))))
active1(app2(X1, X2)) -> app2(active1(X1), X2)
active1(app2(X1, X2)) -> app2(X1, active1(X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
active1(s1(X)) -> s1(active1(X))
active1(zWadr2(X1, X2)) -> zWadr2(active1(X1), X2)
active1(zWadr2(X1, X2)) -> zWadr2(X1, active1(X2))
active1(prefix1(X)) -> prefix1(active1(X))
app2(mark1(X1), X2) -> mark1(app2(X1, X2))
app2(X1, mark1(X2)) -> mark1(app2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
s1(mark1(X)) -> mark1(s1(X))
zWadr2(mark1(X1), X2) -> mark1(zWadr2(X1, X2))
zWadr2(X1, mark1(X2)) -> mark1(zWadr2(X1, X2))
prefix1(mark1(X)) -> mark1(prefix1(X))
proper1(app2(X1, X2)) -> app2(proper1(X1), proper1(X2))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(s1(X)) -> s1(proper1(X))
proper1(zWadr2(X1, X2)) -> zWadr2(proper1(X1), proper1(X2))
proper1(prefix1(X)) -> prefix1(proper1(X))
app2(ok1(X1), ok1(X2)) -> ok1(app2(X1, X2))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
s1(ok1(X)) -> ok1(s1(X))
zWadr2(ok1(X1), ok1(X2)) -> ok1(zWadr2(X1, X2))
prefix1(ok1(X)) -> ok1(prefix1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

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