For instance, the following seems obvious but I don't know how to state the proof formally (and directly):(adsbygoogle = window.adsbygoogle || []).push({});

Show [tex]A \cap (B-A) = \{\} [/tex]

Here is a try:

For any [tex]x \in U \ if \ x \in A[/tex] then [tex] x \notin (B-A)[/tex]

therefore [tex] A \cap (B-A) = \{\} [/tex]

there is something missing....

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# How do you prove things with null

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