1
$\begingroup$

If I say some real numbers are rational it can be denoted in first order logic,

$(\exists x)$ $(real(x) \land rational(x))$

Where $real(x)$ - x is a real number.

$rational(x)$ - x is a rational number.

But if I say all the computers are fast, it denotes by,

$(\forall x)$ $(computer(x) \to fast(x))$

Where $computer(x) $ - x is a computer.

$fast(x)$ - x is fast.

So I want to know we can use implication in second example because that says "for all"? If there was existential quantifier should I use "and" for "implies"?

Please someone tell me the reason behind using implication there.

$\endgroup$
  • $\begingroup$ Yes; the correct translation needs $\land$ with $\exists$ and $\to$ with $\forall$. $\endgroup$ – Mauro ALLEGRANZA Jun 4 '16 at 12:07
  • $\begingroup$ But there is small problem left. Think a sentence like this. " Every person has at least one secret which is not shared by that person" and if I traslate it to this, $(\forall x)$$(\exists y)$$[person(x) \land secret(y) \to \lnot shared(secret(y),x)$ , is this correct? Can I use implies when there are $\forall$ And $\exists$ Exists? $\endgroup$ – Samitha Nanayakkara Jun 4 '16 at 13:39
0
$\begingroup$

My answer here should clarify the meaning of quantifiers for you, and also help you answer your question here. It is not a matter of anyhow choosing some symbols. The symbols are supposed to make perfect sense!

When you want to say "Some real number is rational", note that it means the same thing as "Something is both a real number and a rational number" (think about it and you will agree), and hence also "Some rational number is real". That is precisely why it corresponds to:

$\exists x\ ( real(x) \land rational(x) )$.

Similarly, when you want to say "Every computer is fast", it means the same as "Given anything, if it is a computer then it is fast". Why? Again you can simply think through it to convince yourself that they are the same, but for this case I want to add a note. You see, the former concerns only computers, right? That means that you don't care about non-computers. Now the latter is on the surface about everything, but the inner conditional statement only concerns computers too! So it is:

$\forall x\ ( computer(x) \rightarrow fast(x) )$.

Got it?

For your question hiding in your comment about:

Every person has at least one secret which is not shared by that person.

Your attempt is incorrect. You clearly intend "$secret$" to be a property, so "$secret(y)$" is a true-false statement about $y$. So "$shared(secret(y),x)$" makes no sense. Instead you would want "$sharedby(y,x)$". (I've also changed the name to better reflect the intended meaning.) Furthermore, it is technically not right to put the quantifiers all at the front. They are supposed to reflect the structure of what you want to assert!

For any person $x$:

  There is some secret $y$:

    $y$ is not shared by $x$.

Translate accordingly:

$\forall x\ ( person(x) \rightarrow$

$\exists y\ ( secret(y) \land$

    $\cdots$

$)$

$)$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Wow. What a great explanation. Now I understand how we use implication and "and" there. To make it sure if I take this example which is " brothers are siblings" we only concern about brothers right? Also siblings doesn't imply they are brothers. So I have to use $(\forall x)$$[brother(x) \to sibling(x)]$ ? Please correct me if I'm wrong. And also How did you get implication after person(x)? That's not clear. $\endgroup$ – Samitha Nanayakkara Jun 4 '16 at 16:59
  • $\begingroup$ @User9125: Yes to your first question. To your second question it's the same reason. You want to say something about every person. So you need to say "For anything, if it is a person then ...". $\endgroup$ – user21820 Jun 5 '16 at 1:39
  • $\begingroup$ @User9125: A minor point is that your English sentence actually conveys a slightly more specific thing when you say it, namely: "$\forall x\ ( brotherof(x,me) \rightarrow siblingof(x,me) )$". $\endgroup$ – user21820 Jun 5 '16 at 1:40
  • 1
    $\begingroup$ Thank you very much. That makes more sense :) $\endgroup$ – Samitha Nanayakkara Jun 5 '16 at 2:12

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.