"Everyone who Patricia likes, Sue doesn't like"
Let $L(x,y)$ stand for "$x$ likes $y$" and $p,s$ for Patricia and Sue, respectively.

Then the statement in logic is:

$\forall x (L(p,x) \implies \neg L(s,x))$

$= \forall x(\neg L((p,x) \vee \neg L(s,x) )$
$=\forall x \neg(L(p,x)\wedge L(s,x))$
$= \neg \exists(L(p,x) \wedge L(s,x))$

I don't understand the first statement of equivalence, $\forall x(\neg L((p,x) \vee \neg L(s,x) )$, and I also don't understand why the original statement in logic is valid, since:

$\neg P \vee \neg Q = \neg(P \wedge Q)$, and a true hypothesis cannot imply a false conclusion: $P \implies \neg Q$ is invalid, right? since the conclusion must be true if the hypothesis (premise) is true, in other words, the conclusion is "forced" to be true by the premise, right?

So if $P \implies \neg Q$ is invalid, then how can the original statement, $\forall x (L(p,x) \implies \neg L(s,x))$, which is the form of an invalid argument, be valid? $P$ is true, yet $Q$ is false. And how is the original statement in logic equivalent to the first equivalence, $\forall x(\neg L((p,x) \vee \neg L(s,x) )$?

  • 1
    $\begingroup$ You are working in predicate logic and not in propositional calculus ... Thus when you translate the statement "Everyone who Patricia likes, Sue doesn't like" as a propositional formula, you get $P \to \lnot Q$ which (as you say correctly) is not valid (it is not a tautology). Reading it as a predicate calculus formula you make a "finer" logical analysis of its structure, getting : $\forall x (L(p,x) \to \lnot L(s,x))$, but neither this formula is valid. What you have to prove is the equivalence between this formula and $∀x(¬L((p,x)∨¬L(s,x))$. 1/2 $\endgroup$ – Mauro ALLEGRANZA Apr 13 '15 at 7:33
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    $\begingroup$ Having poved that $∀x(L(p,x) \to \lnot L(s,x)) \equiv ∀x(¬L((p,x) ∨ ¬L(s,x))$, now we can conclude that the formula : $∀x(L(p,x)→¬L(s,x)) \leftrightarrow ∀x(\lnot L((p,x) ∨ \lnot L(s,x))$ is valid. 2/2 $\endgroup$ – Mauro ALLEGRANZA Apr 13 '15 at 7:35
  • $\begingroup$ @MauroALLEGRANZA thank you so much. I would accept this as the final answer and upvote it if you type this in the answer section. $\endgroup$ – Emi Matro Apr 14 '15 at 19:16

The original sentence is: "Everyone who Patricia likes, Sue does not like."

The first statement in logic reads: "For every person $X$ in the set the following is true: if $X$ is liked by Patricia, then $X$ is not liked by Sue."

The next statement in logic reads: "For every person $X$ in the set the following is true: $X$ is not liked by Patricia, or not liked by Sue, or not liked by both Patricia and Sue."

Clearly all three statements are equivalent.

  • $\begingroup$ thanks +1 So then $P$ is $L(p,x)$ and Q is $\neg L(s,x)$? I just don't understand why the premise leads to a negation...isn't that invalid? $\endgroup$ – Emi Matro Apr 13 '15 at 2:05
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    $\begingroup$ No, that is not invalid. In logic the NOT-operator is a key ingredient in formulating statements. This allows one to say: "X is liked by Patricia" implies "X is not liked by Sue". There is nothing wrong with that. $\endgroup$ – M. Wind Apr 13 '15 at 3:27

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