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I want to translate this sentence

There exists $a$ such that if for all $b$ different from $a$, $b$ has the propriety $P$ then $a$ has the propriety $Q$

I translated it like this :

$\exists a. [(\forall b. b \neq a \implies P(b)) \implies Q(a)] $

but it looks weird.

(mostly because if I have the sentence :

There exists $a$ such that if for all $b$ different from $a$, $b$ has the propriety $P$ then a has the propriety $\neg P$

I would translate it like this :

$\exists a. [(\forall b. b \neq a \implies P(b)) \implies \neg P(a)] $

but it can be read as

$\exists a. \neg P(a)$

)

Am I doing this right or are there mistakes I don't see ?

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It indeed looks weird, and that's because typically an existential goes hand in hand with a conjunction ($\land$), rather than a conditional ($\rightarrow$).

But: 'typically' is not the same as 'always'. Indeed, I would say this case is one of those rare exceptions where you do use a conditional. Or at least, your translation coincides exactly with my interpretation of the English sentence as well.

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  • $\begingroup$ Yep, that's why it troubles me. And I guess you don't find a simpler way to translate it? $\endgroup$ – Lhooq Jan 13 at 9:13
  • $\begingroup$ @Lhooq I figured. But again, it's not a hard rule that any existential has to come with a $\land$ .. it's just that many naturally occurring statements translate that way. But in this case, a conditional is the right operator. And I would translate it exactly the way you did. $\endgroup$ – Bram28 Jan 13 at 15:58
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I think your brackets are in the wrong place. Should be

$\exists a. [\forall b. b \neq a \implies (P(b) \implies Q(a))]$

Alternatively you could write this as:

$\exists a. [\forall b. (b = a) \lor (P(b) \implies Q(a))]$

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  • $\begingroup$ What I want to say is that if for all $b$ different from $a$, $P(b)$, then $Q(a)$. From what you wrote I read "for all $b$ different from $a$, if there's one $b$ such that $P(b)$ then $Q(a)$" which is not exactly the same. And I have some trouble to understand why you put parenthesis around $P(b)$ $\endgroup$ – Lhooq Jan 11 at 13:11
  • $\begingroup$ @Lhooq Sorry, my mistake, the brackets were meant to go around $P(b) \implies Q(a)$. I have fixed this in my answer. $\endgroup$ – gandalf61 Jan 11 at 13:24
  • $\begingroup$ But your formulas don't mean that if for all $b$ different from $a$, $P(b)$ is true then $Q(a)$ is true. Their meaning is "for all $b$ different from $a$, if $P(b)$ is true then $Q(a)$ is true. So if I'm not wrong your formula is larger than mine because mine states that I need all $b$ to be $P(b)$ when yours state that you just need one, no? $\endgroup$ – Lhooq Jan 11 at 13:36
  • $\begingroup$ @Lhooq I think we are interpreting the original English sentence in two different ways. Which is not surprising since English is ambiguous. $\endgroup$ – gandalf61 Jan 11 at 15:37

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