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I've been given the example question...

We can derive the new law A ⊆ A ∪ B as follows:

  • By Associativity and Idempotence, A ∪ (A ∪ B) = A ∪ B.
  • Letting X = A and Y = A ∪ B, this says that X ∪ Y = Y.
  • By the above, this means that X ⊆ Y; that is, that A ⊆ A ∪ B.

I don't understand where "A ∪ (A ∪ B)" originally comes from?

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    It's made up by the person writing the proof. It's designed so that the final conclusion will follow. – 2017-01-09
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    As for why $A\cup (A\cup B)$ is true, note that by associativity $A\cup (A\cup B)=(A\cup A)\cup B$ and by idempotence $(A\cup A)\cup B=A\cup B$ – 2017-01-09
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    @michael-burr How did the person make it up? What is the relevance to "A ⊆ A ∪ B"? – 2017-01-09
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    Work backwards, what do $X$ and $Y$ need to be in order to conclude the desired result? – 2017-01-09

2 Answers 2

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Two sets, $C$ and $D$, are equal iff $C \subseteq D$ AND $D \subseteq C$. Also, $C \subseteq D$ (and vice versa) iff every element of $C$ is an element of $D$. You can use these properties to prove that $A \cup (A \cup B) = A \cup B$.

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The goal is to prove that $A\subseteq A\cup B$.

Now, consider the statement that if $X\cup Y=Y$, then $X\subseteq Y$. Working backwards, in order to get $A\subseteq A\cup B$ as the conclusion, you must use $X=A$ and $Y=A\cup B$.

Now, we must check the hypothesis of the statement above, in other words, one must show that $X\cup Y=Y$ or that $A\cup(A\cup B)=A\cup B$. However, this is true as you state.

Therefore, since $X$ and $Y$ are arbitrary in the statement above, using the special case where $X=A$ and $Y=A\cup B$, we get exactly the desired conclusion. Therefore, we choose the $X=A$ and $Y=A\cup B$ in order to engineer that the desired conclusion occurs - $X$ and $Y$ are variables and we can choose them however we wish.