# Predicate Calculus statement

I've been asked to write down a statement using predicate calculus and it is confusing me a great deal.

I've got statement A "no dog can fly" and B "There is a dog which can fly" D = set of all dogs , F = set of all creatures that can fly P(x) is the proposition that "creature x can fly" Q(x) is the proposition that "creature x is a dog"

How do I write statements A and B using predicate calculus in terms of P(x)?

I wrote for A: ∀x(P(x)→¬Q(x) and B: ∃x(P(x)→Q(x) but this doesn't seem right to me at all. Anyone got a suggestion?

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## 1 Answer

The statement A is OK, apart from a missing parenthesis. Statement B should be something like $\exists x(Q(x)\land P(x))$.

Your version of B would be true if there were, for example, no flying creatures.

There are always many equivalent ways of stating things. Closer in tone to the English statement of A is $\forall x(Q(x)\longrightarrow \lnot P(x))$. Or maybe $\lnot\exists x(Q(x)\land P(x))$.

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Ah yes, darn it I should have thought of the intersection symbol. That makes more sense. – user2661167 Aug 31 '13 at 3:29
If i have to write the statements with regards to Q(x) then would it be this? A) ∀x((Q(x)→ ¬P(x)) B) ∃x(Q(x) ^ P(x)) ? – user2661167 Aug 31 '13 at 3:36
I don't know what you mean by with regards to $Q(x)$. Your original version of A was correct, and so is the one in the comment. Your B in the comment is correct. I had mentioned these things in my answer above. – André Nicolas Aug 31 '13 at 3:48
hmmm ok, the question asked me to declare with regards to P(x) then Q(x), its got me a bit confused – user2661167 Aug 31 '13 at 3:56
I do not know what "declare with respect to" might mean. For A, we have two equivalent statements, one with $Q(x)\longrightarrow \lnot P(x)$ and the other with $P(x)\longrightarrow \lnot Q(x)$. The first might "declare with regards" to $Q(x)$, the second with regards to $P(x)$. – André Nicolas Aug 31 '13 at 4:01