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(i) (A ∪ C) - B ⊆ (A - B) ∪ C

(ii) (A ∩ C) - B = (A - B) ∩ C

I was able to find the solution to part (ii) by finding the truth values for (A ∩ C) - B and (A - B) ∩ C in their respective columns then I was able to conclude that these two statements are equal since the truth values for each statement was the same. However, I am having difficulty figuring out how to show one statement is the subset of the other using truth tables and their corresponding truth values. In other words, how do you show (A ∩ C) - B is a subset of (A - B) ∪ C

Any hints would be greatly appreciated!

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    The truth table of $(A-B) \cup C$ should be true whenever the truth table of $(A \cup C) - B$ is true. – 2017-01-31

1 Answers 1

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The subset maps to the conditional.

$X\subseteq Y$ iff $\forall z~[~z\in X\to z\in Y~]$

Which, in a truth table representation, means that whenever $\color{silver}{z\in}X$ is true then so too must $\color{silver}{z\in}Y$ be true.

That is, that $(A \cup C) - B \subseteq (A - B) \cup C$ is true if $~((a\vee c)\wedge\neg b)~\to~((a\wedge\neg b )\vee c~$ is a tautology.