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

(0) Obligation:

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

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

APP(cons(X, XS), YS) → ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
PREFIX(L) → NIL
PREFIX(L) → PREFIX(L)
ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

PREFIX(L) → PREFIX(L)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

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

(6) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [ ] on the rule
PREFIX(x0)[ ]ⁿ[ ] → PREFIX(x0)[ ]ⁿ[ ]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Instantiation
PREFIX(L)[ ]ⁿ[ ] → PREFIX(L)[ ]ⁿ[ ]
    by OriginalRule from TRS P

(7) NO

(8) Obligation:

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

ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
APP(cons(X, XS), YS) → ACTIVATE(XS)
ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)

The TRS R consists of the following rules:

app(nil, YS) → YS
app(cons(X, XS), YS) → cons(X, n__app(activate(XS), YS))
from(X) → cons(X, n__from(s(X)))
zWadr(nil, YS) → nil
zWadr(XS, nil) → nil
zWadr(cons(X, XS), cons(Y, YS)) → cons(app(Y, cons(X, n__nil)), n__zWadr(activate(XS), activate(YS)))
prefix(L) → cons(nil, n__zWadr(L, prefix(L)))
app(X1, X2) → n__app(X1, X2)
from(X) → n__from(X)
nil → n__nil
zWadr(X1, X2) → n__zWadr(X1, X2)
activate(n__app(X1, X2)) → app(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__zWadr(X1, X2)) → zWadr(X1, X2)
activate(X) → X

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• APP(cons(X, XS), YS) → ACTIVATE(XS)
  The graph contains the following edges 1 > 1

• ACTIVATE(n__app(X1, X2)) → APP(X1, X2)
  The graph contains the following edges 1 > 1, 1 > 2

• ACTIVATE(n__zWadr(X1, X2)) → ZWADR(X1, X2)
  The graph contains the following edges 1 > 1, 1 > 2

• ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, n__nil))
  The graph contains the following edges 2 > 1

• ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
  The graph contains the following edges 1 > 1

• ZWADR(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
  The graph contains the following edges 2 > 1

(10) YES