For this entire question, please use the following propositional function: $P(x,y)$: $x$ has sent a postcard to $y$. Translate the following quantified propositions to English sentences. Try to use sentences as natural as possible.
(a) $\forall x\exists y \lnot P(x, y)$
(b) $\exists x\forall y \lnot P(x, y)$
(c) $\forall x\exists y \lnot P(y, x)$
(d) $\exists x\forall y \lnot P(y, x)$
(a) Every $x$ has some $y$ to whom he has not sent a postcard.
(b) Some $x$ has not sent a postcard to every $y$.
(c) Every $x$ has some $y$ from whom he has not received a postcard.
(d) Some $x$ has not received a postcard from every $y$.
Do you think my answers are correct?
What language does the expression
belong to? If anything, it belongs to that logic-English mix which some call Loglish. Using Loglish is a very useful step on the way towards producing a translation into natural language (see my Introduction to Formal Logic, §24.1 for lots of examples of Loglish being used this way). So I'm all for using Loglish. But you can't stop there. For Loglish certainly isn't English. Just try saying the displayed expression to any native speaker outside the logic classroom and see what reaction you get!
You are being asked to translate into natural English, and that means go on to drop the $x$s and $y$s -- which, at a first step, typically involves replacing them by natural-language pronouns. You make a rather fumbled first step when you write
where you've replaced the second '$x$' with 'he', and the 'to $y$' with 'to whom'. But you shouldn't do this piecemeal, ditching some of the $x$s say and not the others: it won't even be good LogLish! You need to lose all the related $x$s in one go. And then all the $y$s. That way you will end up, as you are asked to do, with natural English, without any $x$s and $y$s.