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I need to translate 2 sentences into predicate logic using the following definitions:-

A(x, y): x admires y
B(x, y): x attended y
P(x): x is a professor
S(x): x is a student
L(x): x is a lecture
m: mary

Now I want to translate the following 2 sentences into predicate logic:

  1. No student attended every lecture.
  2. No lecture was attended by every student.

My translations:

  1. $ \neg \exists x \space (S(x) \land (\forall y \space L(y) \rightarrow B(x, y) ) ) $
  2. $ \neg \exists x \space (L(x) \land (\forall y \space S(y) \rightarrow B(y, x) ) ) $

Problem:

  1. First of all, I'm not sure if my translations are correct(although I'm convinced that they are).
  2. Second, I am not able to deduce why the following translations are wrong:
    1. $ \neg \exists x \forall y \space (S(x) \land L(y) \rightarrow B(x, y)) $
    2. $ \neg \exists x \forall y \space (L(x) \land S(y) \rightarrow B(y, x)) $
    or,
    1. $ \neg \exists x \space (S(x) \rightarrow (\forall y L(y) \rightarrow B(x, y))) $
    2. $ \neg \exists x \space (L(x) \rightarrow (\forall y S(y) \rightarrow B(y, x))) $

    I know that they are all different, that much is evident by converting each translation into clause form and comparing. But I'm not able to put a finger on what is wrong.
    For the last set of translations, the only reasoning I'm able to use is by translating the predicate into a non-English language and finding that the sentence constructed is not very idiomatic.

  3. Lastly, as a result of 3, I find that using an implication in $\exists$ just seems wrong(as I said, the sentence formed are not very idiomatic when translated into a non-English language(I tried 2 other languages)). Are there any cases where an implication makes sense in an $\exists$.
Likhit
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1 Answers1

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(a) The bracketing is wrong, for a start.

Take the first example in stages, using LogLish (a mix of English and logical symbolism) as a halfway house!

How would you render "x attended every lecture"? $\forall y(Ly \to Bxy)$.

How would you render 'No student is $\varphi$" using a negated existential? $\neg\exists x(Sx \land \varphi x)$

Now you want the case where $\varphi x = x$-attended-every-lecture = $\forall y(Ly \to Bxy)$. So plugging that in, we get

$\neg\exists x(Sx \land \forall y(Ly \to Bxy))$

Do you see why the $Bxy$ needs to be inside the scope of the existential quantifier?

The second example is similar.

(b) On the general issue of translating restricted quantifiers, see e.g. Restricted quantifiers - Logic (and for more, see the reference to Teller's Primer given there).

Community
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Peter Smith
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  • Sorry about that, I took the brackets to be implicit since that's the convention we are using in class(and in the book we are following). The scope of the operators probably looks more obvious on paper than on the screen. I edited the question and added the brackets. – Likhit May 22 '13 at 16:31
  • I read your answer on the Restricted quantifiers link. This brings me to the third question I asked. Does it ever make sense in practical predicate logic to use implication in the existential quantifier? Also, why and how are the quantifiers "restricted"? – Likhit May 22 '13 at 16:38
  • The bracketing is still wrong, given the usual convention that $\land$ binds tighter than $\to$. – Peter Smith May 22 '13 at 16:56
  • I think you are talking about ¬∃x∀y (S(x)∧L(y)→B(x,y)), here the \land is meant to bind tighter first. My question asks why it i.e ¬∃x∀y ((S(x)∧L(y))→B(x,y)) is wrong? – Likhit May 22 '13 at 17:04