8
$\begingroup$

I'm following my first logic course as part of my pre-masters programme. Currently I'm working on predicate logic.

I know that $∀$ is the universal quantifier, which stands for "all" or "every", and $∃$ is the existential quantifier, which stands for "some" or "there is".

In my textbook I tried the following question:

Translate the following sentences into predicate logical formulas. Assume the domain of discourse is human beings. Not all girls love themselves

I used $G$ for "Girl" and $L$ "Loves". My translation was as follows: $$ ¬∀x(Gx \to Lxx) $$ But, the solution the textbook gives is: $$ ∃x(Gx ∧ ¬Lxx) $$

I'm really wondering if both solutions are correct. Actually, I think mine is more precise considering that "Not All" is $¬∀x$ and $∃x$ is "Some".

But I guess I am missing something, or is this just a style thing and are both correct?

$\endgroup$
6
$\begingroup$

The two answers are equivalent.

"$\lnot \forall$" is the same as "$\exists \lnot$".

If not all cats are black, there must be some cat that is not black.

Thus, we have that $$ ¬∀x \ (Gx \to Lxx) \iff ∃x \ ¬(Gx \to Lxx) \text{.} $$

Now we apply the tautological equivalence $$ \lnot (p \to q) \iff (p \land \lnot q) $$

(We can check it with a truth-table: $\lnot (p \to q)$ is TRUE just in case when $p$ is TRUE and $q$ is FALSE.) to get the final result, $$ ∃x \ (Gx \land \lnot Lxx) \text{.} $$

$\endgroup$
  • 2
    $\begingroup$ Thank you for your anwser! However i'm not sure what you mean with the other ingredient. I know that ¬(p→q) and (p∧¬q) are equivalent to each other (exact same truth-table) $\endgroup$ – Bart Oct 9 '17 at 8:16
  • 1
    $\begingroup$ @Bartvh The author was likely showing the full reason why the two are equal for future readers. Even though you only asked about "not all" vs "some", this makes the answer useful for others who might not be as knowledgeable. $\endgroup$ – Brian J Oct 9 '17 at 13:58
2
$\begingroup$

Both are correct. They are equivalent.

$\neg \forall x~(Gx\to Lxx)$ "Not all girls love themselves."

$\exists x~(Gx\wedge \neg Lxx)$ "Some girls don't love themselves."

$\endgroup$
1
$\begingroup$

I would say that the author of the textbook took a wrong approach. The task of translation is to communicate meaning as close as possible to the original. This is especially important in mathematics. Your answer follows English text verbatim, and the author’s answer is transformed. Albeit the transformation is legal, I bet that it baffles readers because transformation of logical formulas does not belong to the translation process. I wholeheartedly agree that your answer is more precise.

If you are picky, then you may want to know that there are circumstances where your answer and the author’s answer are not equivalent. Their equivalence is impossible to prove in intuitionistic logic. Intuitionistic logic is slightly weaker than classical logic that your textbook teaches. Intuitionistic logic does not contain the law of excluded middle, and truth tables are inapplicable to it; statements are proved by inference.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.