Re: Embeddings and negation

Yikes, I meant "But then A |= ~P[a1,...,an] by (2), contradicting our
hypothesis." rather than "But then A |= ~P[a1,...,an] by (3.17),
contradicting our
hypothesis".

On Jan 17, 1:23 pm, "Michael De" <mike...@xxxxxxxxx> wrote:
I'm reading Hodges' "A Shorter Model Theory" and have a question about
a result on pp. 14 (3.17).

Let f: A -> B be an embedding from dom(A) into dom(B). Then for any
atomic formula P(x1,...,xn), A |= P[a1,...,an] iff B |=
P[f(a1),...,f(an)], where each ai is an element of dom(A). It follows
that

(1) A |= P[a1,..,an] doesn't hold iff B |= P[f(a1),...,f(an)] doesn't
hold.

Now suppose moreover that A and B are closed under negation such that

(2) A |= ~P[a1,...,an] iff not A |= P[a1,...,an] (where P is still
atomic).

Let Q be a literal (i.e. an atom or its negation). Then we have

(3.17) If A |= Q[a1,...,an] then B |= Q[f(a1),...,f(an)].

My question is: Does a biconditional version of (3.17) hold? The
statement of (3.17) implies that it does not, otherwise why not state
it as a biconditional? It seems, however, as if it must hold according
to the following reasoning.

Let f and P be as before. Assume B |= ~P[f(a1),...,f(an)] but not A |=
~P[a1,...,an], and the hence the failure of the converse of (3.17).
Then by (2) not B |= P[f(a1),...,f(an)] and hence not A |= P[a1,...,an]
by (1). But then A |= ~P[a1,...,an] by (3.17), contradicting our
hypothesis.

(2) and (3.17) are ok, so I guess my (1) must be incorrect. But suppose
it doesn't hold so that either A |= P and not B |= P or not A |= P and
B |= P, for some atomic P. But that's impossible by the definition of
an embedding, viz., that A |= P iff B |=P.

So.....?

.