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Is the following deduction correct?

$$\begin{align} x\notin X\setminus\{y\}\quad & \iff\quad x\notin X \cap \overline{\{y\}}\\ \\ & \iff\quad x \in \overline{X\cap \overline{\{y\}}}\\ \\ & \iff\quad x \in \overline X \cup \{y\}\\ \\ & \iff\quad x \notin X\text{ or } x =y\end{align}$$

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    $\begingroup$ Yup. ${}{}{}{}$ $\endgroup$ – Clive Newstead Sep 18 '13 at 13:22
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    $\begingroup$ I would strongly advise against using $\overline{X}$ to denote the complement - it denotes the closure - but apart from that, I see nothing to criticize. $\endgroup$ – Daniel Fischer Sep 18 '13 at 13:22
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    $\begingroup$ What about the obvious $x\in X\setminus\{y\}\iff x\in X \land x\neq y$? $\endgroup$ – Michael Hoppe Sep 18 '13 at 13:50
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Yes, the deduction is correct, but it's important to clarify what you're doing and why it holds (particularly when submitting for homework, and/or writing an exam!)

So I'd suggest adding more words, if only to add justification. For e.g., you use DeMorgan's when going from the third to fourth line: $$\quad x \in \overline{X\cap \overline{\{y\}}} \iff x \in \overline X \cup \{y\}\tag{DeMorgan's}$$ Say so!

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What you did is fine but I'd negate both sides and then it's obvious by definition: $$ x\in X\setminus\{y\} \iff x\in X \text{ and } x\ne y $$

Of course, De Morgan's law is present in the form not (P or Q) is the same as (not P) and (not Q).

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