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

(0) Obligation:

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

app(nil, YS) → YS
app(cons(X), YS) → cons(X)
from(X) → cons(X)
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X), cons(Y)) → cons(app(Y, cons(X)))
prefix(L) → cons(nil)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Lexicographic path order with status [LPO].
Precedence:

from₁ > cons₁ > app₂
zWadr₂ > nil > app₂
zWadr₂ > cons₁ > app₂
prefix₁ > nil > app₂
prefix₁ > cons₁ > app₂

Status:

app₂: [1,2]
nil: []
cons₁: [1]
from₁: [1]
zWadr₂: [1,2]
prefix₁: [1]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

app(nil, YS) → YS
app(cons(X), YS) → cons(X)
from(X) → cons(X)
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X), cons(Y)) → cons(app(Y, cons(X)))
prefix(L) → cons(nil)

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE