Re: Proto-Romance *ve:ci:nus 'neighbour'

--- In cybalist@yahoogroups.com, "m_iacomi" <m_iacomi@...> wrote:

> --- In cybalist@yahoogroups.com, "Richard Wordingham" wrote:
>
> >> But vi:ci:nus= vec^in, vi:ci:nita:s= vec^inãtate which shows
> >> an "e" from an long latin /i/.
> >> The posibility should be that the word is not inherited.
> >
> > I apologise if I've already raised this topic.
> >
> > Spanish _vecino_ and French _voisin_ point to *ve:ci:nus (or
> > *vici:nus).
>
> Why?! We deal with unstressed vocalism. Cf. Varvaro - "Linguistica
> Romanza", for instance: "nel sistema 'romanzo comune' [Latin] i
> [short] e: e [short] hanno dato /e/ [...] nel sistema balcanico
> la serie anteriore è la stessa". So "vici:num" (short unstressed
> /i/ only) could explain Spanish <vecino>; OTOH, French <voisin>
> and OF <veisin> are usually explained through VL "*veci:nus".

> > Although Italian _vicino_ appears to derive from _vi:ci:nus_,
>
> If you start up with "vi:ci:nus" (which is CL), you'll end up
> with Romanian "*vicin" which could arguably dissimilate into
> modern "vecin" (Rohlfs admits it too for Italian). Conversely,
> you may try "vici:nus" or "*veci:nus" (as for French); unstressed
> Latin /e/ does not go always in /&/ even in the first syllable:
> "venenum" > "venin", "venire" > "veni", "berbecem" (CL "vervex") >
> "berbec(e)", etc.

You seem to be making the point that all the forms could derive from
any of *veci:nus, *ve:cinus and *vici:nus. You argue that Romanian
_vecin_ might derive from Latin _vi:ci:nus_; why do you boldly
assert that Italian _vicino_ does not (as opposed to need not)
derive from _vi:ci:nus? From what you say, Rohlfs did not dismiss
it. Information from other dialects?

The cleanest explanation is to derive all forms from a single proto-
form. I can explain *ve:ci:nus (derivative from attested by-form
_ve:cus_ of _vi:cus_). *vici:nus would be a dissimilatory
shortening. Where would *veci:nus come from? Or is <e> simply a
convenient notation for the reflex of unstressed /e/, /e:/ and /i/?

Richard.