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I came across following problem

"Every intelligent student is not honest."

And I have to convert this in quantifiers. Straight conversion will be:

∀x [(S(x)∧I(x)) → ¬H(x)] ...(i)

However the solution is given in existential quantifier as follows:

∃x [S(x)∧I(x)∧¬H(x)] ...(ii)

with explanation "There exist intelligent students who are not honest"

Though this sounds and looks correct, how can I convert (i) to (ii) mathematically I mean without verbal interpretation, may be by double negation?

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  • $\begingroup$ What if there are no students? $\endgroup$ Jan 21 '15 at 17:59
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    $\begingroup$ I think the given solution is incorrect. It appears to conflate “Every intelligent student is not honest” and “Not every intelligent student is honest”. $\endgroup$
    – MJD
    Jan 21 '15 at 18:01
  • $\begingroup$ The two sentences are not equivalent. Are you sure you're not misunderstanding what the problem asks for? $\endgroup$ Jan 21 '15 at 18:02
  • $\begingroup$ What are the dots standing in for? Is that the whole sentence? $\endgroup$
    – GFauxPas
    Jan 21 '15 at 18:11
  • $\begingroup$ @GFauxPas those are just "refer this as" (i) $\endgroup$
    – Maha
    Jan 21 '15 at 18:11
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As per the above comments, the given solution is not correct.

From :

$∀x [(S(x) \land I(x)) \rightarrow \lnot H(x)]$

we have to start using the equivalence between $\forall$ and $\lnot \exists \lnot$ to get

$\lnot \exists x \lnot [(S(x) \land I(x)) \rightarrow \lnot H(x)]$.

Then, we have to apply the tautological equivalence between : $\lnot (p \rightarrow \lnot q)$ and $(p \land q)$ [you can check it with a truth-table] and convert the above formula into :

$\lnot \exists x [S(x) \land I(x) \land H(x)]$.

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  • $\begingroup$ can we get this by double negation: ¬¬∀x[(S(x)∧I(x))→¬H(x)]¬∃x¬[(S(x)∧I(x))→¬H(x)]¬∃x[(S(x)∧I(x))∧H(x)] (Since ¬(p→q)≡p∧¬q) ≡ ¬∃x[S(x)∧I(x)∧H(x)] $\endgroup$
    – Maha
    Jan 21 '15 at 20:25
  • $\begingroup$ @awellwisher - yes : $\lnot \forall$ is $\exists \lnot$; thus $\lnot \lnot \forall$ is $\lnot \exists \lnot$. $\endgroup$ Jan 21 '15 at 20:34

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