Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble proving the theorem below:

$A \cup (B - C) = (A \cup B) - (C - A)$

Using the direct method I am to assume that the hypothesis is true. So that is where I start:

$\begin{array}{ccl} A \cup (B-C) & = & \{x \;|\; x\in A \vee x \in (B-C) \} \\ & = & \{x \;|\; x \in A \vee (x \in B \wedge x \not\in C)\} \\ & = & \{x \;|\; (x \in A \vee x \in B) \wedge (x \in A \vee x \not\in C)\} \end{array}$

But I don't know where to go from here. I have $(A \cup B)$ on the right side but how do I derive "$- (C - A)$"?

share|cite|improve this question

1 Answer 1

up vote 1 down vote accepted

Use De Morgan's laws to rewrite $x\in A \lor x\not\in C$ to $\neg(x\not\in A\land x\in C)$, and then commutativity of $\land$ for $\neg(x\in C \land x\not\in A)$.

By the way, it is also useful to be able to do these manipulation at the set level, without explicitly unfolding to a comprehension formula: $$A\cup(B\setminus C)=A\cup(B\cap \overline C)=(A\cup B)\cap(A\cup\overline C)=\ldots$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.