(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) 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.
(10) 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.