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I want to prove that $(A\cap B)\cup(A\cap C) \subseteq (A\cap(B\cup C))$. I noticed that I recently, I have just been applying the laws of logical equivalence (ie: distributivity/commutativity/demorgan's etc). The proof I used was as follows and I wanted to check if this was a valid way because it differed from the solutions.

Proof: Assume $x \in (A\cap B)\cup(A \cap C) \subseteq (A \cap (B \cup C))$

Then $x \in (A \cap B) \vee x \in (A \cap C)$

$ \implies (x \in A \wedge x \in B) \vee (x \in A \wedge x \in C)$

$\implies $ by distributivity of logical statements, $x \in A \wedge (x \in B \vee x \in C)$

Then, $(x \in A) \wedge (x \in B\cup C)$

Thus, $x \in A \cap (B \cup C)$

$\therefore (A\cap B)\cup(A\cap C) \subseteq (A\cap(B\cup C))$

Would this be a valid way as opposed to considering the cases of $x \in (A \cap B)$ and then considering $x \in (A\cap C)$, which was the way the solutions outlined.

Advice would be appreciated.

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  • $\begingroup$ Looks good to me. $\endgroup$ – vadim123 Jun 4 '16 at 21:33
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Your proof is correct. In fact, it has two virtues over the case-proof: 1) it is shorter, and 2) it preserves generality. By the latter I mean that when one uses only equivalence rules, the argument employed may be reversed so to get the converse of the theorem. Thus you can affirm that $A\cap(B\cup C)\subseteq (A\cap B)\cup(A\cap C)$ too.

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