0
$\begingroup$

This question is from Rosen:

If L(x,y): x loves y.

Use quantifiers to express:

"There is someone who loves no one besides himself or herself"

The answer given by textbook is ∃x∀y(L(x,y)↔x=y)

What I don't understand is what does the statement mean when both p and q in p <=> q are false. The statement is true, according to biconditional truth table, but what does the statement actually mean?

My answer for the question was ∃x∀y(L(x,x) ∧ ¬L(x,y) ∧ ¬(x=y))

$\endgroup$
  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Jun 19 '17 at 12:28
  • $\begingroup$ Fix $x$ as John; we have : $∀y(L(John,y) ↔ John=y)$. It means: " a person loves John iff that person is John himself". $\endgroup$ – Mauro ALLEGRANZA Jun 19 '17 at 12:32
  • $\begingroup$ When $y$ is not John ( $\lnot (John = y)$), he does not love John ($\lnot L(John,y)$). $\endgroup$ – Mauro ALLEGRANZA Jun 19 '17 at 12:33
  • 1
    $\begingroup$ wow! thank you so much! very helpful, quite cleared the haziness in my head. One question, how would you recommend I could improve my understanding of logic? Also, If you'd like I could accept your answer if wrote one. @MauroALLEGRANZA $\endgroup$ – momo Jun 19 '17 at 12:47
  • $\begingroup$ Reading some logical textbook... :-) See e.g. Peter Smith, An Introduction to Formal Logic. $\endgroup$ – Mauro ALLEGRANZA Jun 19 '17 at 12:49
1
$\begingroup$

Consider to set $x$ as $John$ (we "name" him).

We have:

$∀y(L(John,y) ↔ John=y)$,

that means "a person loves John iff that person is John himself".

What happens when $y$ is not $John$ ?

Well: $(John=y)$ is false and he does not love $John$, i.e. $L(John,y)$ is false also, and we know that $p ↔ q$ is true when both $p$ and $q$ are false (they are "equivalent").

What about the proposed:

$∃x∀y(L(x,x)∧¬L(x,y)∧¬(x=y))$ ?

We have that (using again $John$ as $x$): $∀y(L(John,John)∧¬L(John,y)∧¬(John=y))$.

The $∀y$ quantifier means "for all"; thus, instantiating it with $John$, we get:

$L(John,John)∧¬L(John,John)∧¬(John=John)$

that is contradictory and also false (see the part $¬(John=John)$), contrary to our intentions.

$\endgroup$

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.