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Attempting to apply more flexible, informal reasoning to predicate logic as demonstrated helpfully to me by another user in answer to my last question.

$\{x \in B \mid x \notin C\} \in \mathscr P(A)$

I read this as "For every item 'x' that's in the set B, if it's not in the set C, then it's a subset of A."

Seems like this can pretty simply be rewritten as

$\forall x \left( \left(x \in B \land x \notin C \right) \to x \in A \right)$

Is this accurate?

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    $\begingroup$ Your final symbolic expression is correct, but the verbal translation with which you started is not. It should be: If $x$ is in the set $B$ but not in the set $C$, then $x$ is an element of $A$. $\endgroup$ – Brian M. Scott Jun 24 '16 at 0:25
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    $\begingroup$ Yes, the choice of the word "subset" is wrong. The set of all such $x$ is a subset of $A$, which means that each such $x$ is an element of $A$. $\endgroup$ – Thomas Andrews Jun 24 '16 at 0:27
  • $\begingroup$ Ah, that's an important distinction! So far, I've been thinking of membership in a Power Set strictly in terms of subsets, that seems like the sort of thing that could have bitten me not too far down the road. Thanks again Brian! $\endgroup$ – user242007 Jun 24 '16 at 0:27
  • $\begingroup$ It can be even more simply written as $B // C \subset A$. $\endgroup$ – fleablood Jun 24 '16 at 0:44
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    $\begingroup$ Note we can rewrite {x in B| x not in C} as $B \cap C^c$ as $B- C$. (or whatever notation indicates $B$ excluding $C$). $X \in P(A)$ simply means $X \subset A$. So we can simplify $B -C \subset A$. Which way easier for my mushy mind to read. $\endgroup$ – fleablood Jun 24 '16 at 0:48
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As noted in comments, the word "subset" is incorrect in your translation, but your formalized version is correct.

The careful way of writing this out is first to note:

$$\left\{x\in B\mid x\notin C\right\}\in \mathcal P(A)$$

is equivalent to:

$$\left\{x\in B\mid x\notin C\right\}\subseteq A$$

which is equivalent to:

$$\forall x\in \left\{x\in B\mid x\notin C\right\}:x\in A$$

which is equivalent to:

$$\forall x:\left((x\in B\land x\notin C)\implies x\in A\right)$$

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