# Translate a sentence into predicate logic

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 ?

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:

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

• 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 $P(b)$ Jan 11 '19 at 13:11
• @Lhooq Sorry, my mistake, the brackets were meant to go around $P(b) \implies Q(a)$. I have fixed this in my answer. Jan 11 '19 at 13:24
• 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? Jan 11 '19 at 13:36
• @Lhooq I think we are interpreting the original English sentence in two different ways. Which is not surprising since English is ambiguous. Jan 11 '19 at 15:37

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.

• Yep, that's why it troubles me. And I guess you don't find a simpler way to translate it? Jan 13 '19 at 9:13
• @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. Jan 13 '19 at 15:58