Termination w.r.t. Q of the given QTRS could not be shown:

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

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

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

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

APP(cons(X, XS), YS) → APP(XS, YS)
FROM(X) → FROM(s(X))
ZWADR(cons(X, XS), cons(Y, YS)) → APP(Y, cons(X, nil))
ZWADR(cons(X, XS), cons(Y, YS)) → ZWADR(XS, YS)
PREFIX(L) → ZWADR(L, prefix(L))
PREFIX(L) → PREFIX(L)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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

(8) Obligation:

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

APP(cons(X, XS), YS) → APP(XS, YS)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(cons(X, XS), YS) → APP(XS, YS)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

[APP₂, cons₂]

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

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

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

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) TRUE

(13) Obligation:

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

ZWADR(cons(X, XS), cons(Y, YS)) → ZWADR(XS, YS)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ZWADR(cons(X, XS), cons(Y, YS)) → ZWADR(XS, YS)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZWADR(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

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

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

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

The set Q consists of the following terms:

app(nil, x0)
app(cons(x0, x1), x2)
from(x0)
zWadr(nil, x0)
zWadr(x0, nil)
zWadr(cons(x0, x1), cons(x2, x3))
prefix(x0)

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