stumbled across this on my exercise sheet:

was just wondering if anyone could show me how to do (a) so i can attempt the rest?

i've written the answer down as Ex (S(x) => K(x, logic)) but need someone to clarify for me whether i've done this correct. if not, where have i gone wrong?

thanks in advance for any help

# predicate logic question

Page 1 of 1## 4 Replies - 2032 Views - Last Post: 07 May 2013 - 07:15 AM

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**Replies To:** predicate logic question

### #2

## Re: predicate logic question

Posted 05 May 2013 - 05:14 AM

First of all you're missing a quantifier for x. As it stands now, your x comes out of nowhere.

There's another mistake, but it's easier to explain why it's wrong, once you've picked a quantifier, so do that first.

There's another mistake, but it's easier to explain why it's wrong, once you've picked a quantifier, so do that first.

### #3

## Re: predicate logic question

Posted 06 May 2013 - 10:43 PM

Quote

First of all you're missing a quantifier for x. As it stands now, your x comes out of nowhere.

I think he is using "E" as the existential quantifier. Perhaps you are referring to the fact that he does not specify "There exists an x

**in domain**"? Though it may seem sloppy, it is actually valid notation to leave off the domain that x belongs to and just say "There exists x". In such cases, the domain must be implied. In this example, there is only one domain so it is automatically implied.

@Idonknow, the problem with your statement is that it is true if there is a student that knows logic, or if x is not a student. So, even if no students knew logic, your statement would still be true for mary and sue. You need to make a statement that is false if no students know logic.

### #4

## Re: predicate logic question

Posted 07 May 2013 - 06:53 AM

### #5

## Re: predicate logic question

Posted 07 May 2013 - 07:15 AM

Even so, in most cases, you deal with a domain and a co-domain, and explicit is

∃ x ∈ Domain: S(x) => K(x, logic)

would be correct here.

**almost**always better than implicit, so saying that there exists a value in the domain versus co-domain is almost always relevant∃ x ∈ Domain: S(x) => K(x, logic)

would be correct here.

This post has been edited by **Dogstopper**: 07 May 2013 - 07:17 AM

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