0
$\begingroup$

I am revising predicate logic by answering questions from past papers, but haven't been given the answers. So I was hoping to check to make sure I am understanding this area properly.

The question is:

Let the logical statement K(a,b) stand for "a knows b", where the universe of discourse is the set of people at a party hosted by Liam.

Translate the following into logical notation:

(a) Liam knowns everyone at the party.

(b) Everyone knows each other.

(c) There are people who do not know each other.

(d) For any two people who do not know each other there is someone who know them both.

(e) Translate the following into English:

$\exists x \forall y (y \not = x) \rightarrow (K(y,x) \land \neg K(x,y))$

My answers are:

(a) Vb (K(Liam,b))

(b) Va Vb (K(a,b))

(c) Ea Eb (~K(a,b) & ~K(b,a))

(d) Va Vb (~K(a,b) & ~K(b,a) ==> Ey (K(y,a) & K(y,b)))

(e) There is at least one person NOT at the party who doesn't know anyone at the party, but everyone at the party knows them. (English is not my first language, so I might be completely wrong here).

$\endgroup$
  • 1
    $\begingroup$ The universe of discourse is people at the party. There is no way to talk about people not at the party. $\endgroup$ – Derek Elkins Jan 2 '18 at 21:23
  • 1
    $\begingroup$ Your symbolizations for a through d are correct, but like Derek said, for e you are still exclusively talking about people at the party. Also, here is a tutorial for Mathjax so you can typeset those expressions and make them look nice and much more readable (and make everyone much happier! :) ) $\endgroup$ – Bram28 Jan 2 '18 at 21:33
  • $\begingroup$ @Bram28 Thank you for confirming a-d, and I am sorry about the formatting, I will fix it. Do you know what the answer is for e? $\endgroup$ – Shannon Jan 3 '18 at 2:16
  • 1
    $\begingroup$ @Shannon It is basically what you have, but without the NOT. You also may want to explicitly say that the person doesn't know anyone else at the party, but that everyone else does know this person. ... also, I used mathjax for e) ... just so you have some idea how this is done. $\endgroup$ – Bram28 Jan 3 '18 at 2:37
2
$\begingroup$

(a) Yes, you're correct.

$\forall x;K(Liam,x)$

(b) Also correct.

$\forall x,y;K(x,y)$

(c) Also correct.

$\exists x,y;(\neg K(x,y)\land\neg K(y,x))$

(d) Also correct.

$\forall x,y;(\neg K(x,y)\land\neg K(y,x)\implies\exists z;(K(z,z)\land K(z,y)))$

(e) We're given $\exists x;\forall y;(y \neq x\implies K(y,x)\land\neg K(x,y))$

There is at least one person NOT at the party who doesn't know anyone at the party, but everyone at the party knows them.

No. There is no one not at the party. It was given at the beginning that "the universe of discourse is the set of people at a party hosted by Liam".

Parsing roughy into english, the statement says "There exists x such that for all y, y does not equal x implies y knows x and x does not know y".

The "x does not equal y" means that they are not the same person.So let's get rid of that part, and replace "x" with "a person", and "y" with an "other".

"There exists a person such that for all others, the other person knows the first person and the first person does not know the other person."

Now we can see that this basically means

"There is at least one person at the party that everyone else knows, but who doesn't know anyone else."

$\endgroup$

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.