I have attached the image of my membership table. I have supposedly mistook three values I marked with green question marks. Can someone please tell me why they are wrong? the correct values are ones immediately left of the question marks.
1 Answer
$\begingroup$Perhaps not a direct answer to your question, but we have $(A\cup B)\setminus C = (A\cup B)\cap C^c = (A\cap C^c)\cup (B\cap C^c) = (A\setminus C)\cup (B\setminus C)$
So, the two are indeed equivalent.
For a truth table:
$$\begin{array}{c|c|c|c|c|c|c|c} A & B & C & A\cup B & (A\cup B)\setminus C & A\setminus C & B\setminus C & (A\setminus C)\cup (B\setminus C)\\ \hline \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 1 & 1 & 0 & 1 & 1\\ 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0\\ \hline 1 & 0 & 0 & 1 & 1 & 1 & 0 & \color{red}{1}\\ 1 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\ \hline 1 & 1 & 0 & 1 & 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \end{array}$$
That being said, looking between my truth table and yours, you did make a mistake in the final column, for $A=1,B=0,C=0$, you should have had a 1 in the entry (as it is the union of the previous two columns and there is a 1 in the corresponding $A\setminus C$ column), but it is otherwise correct. The conclusion should be that the $(A\cup B)\setminus C$ column should exactly match the $(A\setminus C)\cup (B\setminus C)$ columns, proving their equality.
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