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If we have

A(X) = x is in this arcade

W(X) = "x won a grand prize"

P(x) = "x played Pac-Man"

how would I express "A person in this arcade won a grand prize," and "Nobody in this arcade played Pac-Man." As premises in a quantified logical statement.

What I have done so far is $\forall x(\neg A(x) \implies P(x))$ and $\exists x (A(x) \implies W(x)$

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    $\begingroup$ I don't think that I'm allowed to answer until you've shown work. Place to start would be to consider the various ways of using the tools: $\forall x, ~\exists x,~$ and $~\neg (\cdots)$. From this perspective, what have you tried and where are you having trouble? $\endgroup$ – user2661923 Oct 3 '20 at 18:15
  • $\begingroup$ yeah I edited the original post to show what I think the answer is $\endgroup$ – ej21vf Oct 3 '20 at 18:27
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"A person in this arcade won a grand prize"

$\exists x (A(x) \implies W(x))$

See e.g. here for a detailed explanation of why your formulization is incorrect: As a rule of thumb, with $\exists$ use $\land$ and with $\forall$ use $\implies$, so instead write

$\exists x(A(x) \land W(x))$

"Nobody in this arcade played Pac-Man."

$\forall x(\neg A(x) \implies P(x))$

Your formula says "Everyone who is not in this arcade played Pac-Man". But what you want to say is "Everyone who is in this arcade did not play Pac-Man", that is, the negation should be applied to the playing Pac-Man, not to the being in the arcade:

$\forall x (A(x) \implies \neg P(x))$

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  • $\begingroup$ +1 nice detail.... $\endgroup$ – user2661923 Oct 4 '20 at 9:57
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$[~\exists x ~(A(x) \wedge W(x))~]$
and
$\{\neg ~~[\exists x ~(A(x) \wedge P(x))~]~~\}$.

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