Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx3_3_25_Bor03

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.