Basic Question about Universal Quantifiers in Predicate Logic

Question

so I'm doing some predicate logic and I've stumbled upon a little confusion.

$∀c(R(c) \implies Y(c))$

$\neg Y(z)$

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Therefore: $\neg R(z)$

I don't understand how that works. What I do know, is that if $Y(z)$ is false, then yes, $R(z)$ must also be false, because there is no case where $R(z)$ is true and $Y(z)$ is false. But, there is a "for all c" quantifier there, doesn't that mean that for all inputs, $R(c)$ holds and so it must be true? I don't really understand, I might have read the definition of a quantifier the wrong way.

Help is much appreciated, thanks in advance!

Answer 1

But, there is a "for all c" quantifier there, doesn't that mean that for all inputs, $R(c)$ holds and so it must be true?

You seem to be confused about the meaning of the implication operator. It might help to recall the definition: $A\implies B \equiv \neg A \lor B$. $A\implies B$ is just a statement about the existing truth values of a pair of logical propositions $A$ and $B.$

Applying this definition, we can infer from your two assumptions that $\neg R(z)\lor Y(z)$ is true. Since $Y(z)$ is assumed to be false, we must have $R(z)$ being false, as required.

Answer 2

But, there is a "for all c" quantifier there, doesn't that mean that for all inputs, $R(c)$ holds and so it must be true?

No, for all values of input c, $ R(c) \implies Y(c) $ is true, since that whole thing is surrounded by the parentheses, so the quantifier applies to that whole thing. You have $ \forall c. P $ where $ P $ is "$ R(c) \implies Y(c) $", not "$ R(c) $".

Just like without quantifiers, if $ R(c) $ is true, then $ Y(c) $ must also be true, regardless of the value of $ c $.