[Math] The membership table for (A U B) – C = (A – C) U (B – C)

elementary-set-theory

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.

enter image description here

Best Answer

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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