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

$∀c(R(c) \implies Y(c))$

$\neg Y(z)$


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!

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  • $\begingroup$ The conditional holds for all particular $z$, but conditionals are true whenever the antecedent is false. $\endgroup$ – Malice Vidrine Oct 1 '18 at 23:35
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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.

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  • $\begingroup$ That is not the definition of implication! Just because they are equivalent in classical logic does not imply that implication is defined that way. Even if you choose to do so, you do so only because implication has a certain meaning and you observe that "not A or B" has the same truth-table as what you want implication to have, which of course is based on the prior intended meaning of implication. $\endgroup$ – user21820 Oct 2 '18 at 15:57
  • $\begingroup$ It (or the equivalent $A\implies B \equiv \neg [A\land \neg B]$) is often presented as The Definition of Implication to save time in introductory courses, but it can indeed be derived from other principles of logic. See my formal derivation at dcproof.com/DeriveImplies.html $\endgroup$ – Dan Christensen Oct 2 '18 at 16:09
  • $\begingroup$ You're mistaken. The implication is most often defined in terms of the truth-table, not in terms of boolean equivalences. And I am not interested in your website; it is (still) not conducive for mathematical pedagogy. $\endgroup$ – user21820 Oct 2 '18 at 16:32
  • $\begingroup$ Not coincidentally, the truth tables are identical. And what is a truth table other than a kind of abbreviation leaving out the implicit logical connectors. Interested readers can see my derivation of each line of the truth table for implication (among other things) as my blog posting dcproof.com/IfPigsCanFly.html $\endgroup$ – Dan Christensen Oct 2 '18 at 16:40
  • $\begingroup$ Interested readers should note that in fact that equivalence is only valid for classical logic and breaks down in other well-known logics, precisely for the reason that I stated; the core meaning of implication is captured differently in different logics. It only so happens that in classical logic "A implies B" is equivalent to "not A or B". $\endgroup$ – user21820 Oct 2 '18 at 16:43
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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 $.

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  • $\begingroup$ Right. @user472288, take careful note of "that whole thing is surrounded by the parentheses". For every $c$, if that $c$ satisfies $R$ then it also satisfies $Y$. So for every $c$, if that $c$ does not satisfy $Y$ then can it satisfy $R$? $\endgroup$ – user21820 Oct 2 '18 at 9:17

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