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Can someone please verify whether my proofs are logically correct? :-)

$A\cup \varnothing = A $

Proof: Let $x\in A \cup \varnothing$. Then $x \in A$ or $x \in \varnothing$. If $x \in A$, then $A \subseteq A$. If $x \in \varnothing$, this forms a contradiction (since the empty set is empty). Therefore $x \in A$ and so $A \cup \varnothing \subset A$.

Now let $x \in A$. Then $x \in A \cup \varnothing$ by disjunctive amplification. Then $A \subset A \cup \varnothing$, and so $A\cup \varnothing = A$. $\square$

$A\cap \varnothing = \varnothing$

Proof: Let $x \in A \cap \varnothing$. Then $x \in A$ and $x \in \varnothing$. This forms a contradiction, since the empty set is empty. Therefore, there does not exist such an $x \in A \cap \varnothing$, and so $A\cap \varnothing =\varnothing$. $\square$

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    $\begingroup$ Proof is good and clear! $\endgroup$ – max_zorn Jan 8 '18 at 0:20
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    $\begingroup$ @max_zorn Your statement is false: did you see the "If $x\in A$, then $A\subseteq A$"? $\endgroup$ – user228113 Jan 8 '18 at 0:23
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    $\begingroup$ Still OK, perhaps write instead: If $x\in A$, then clearly $x\in A$ as required." $\endgroup$ – max_zorn Jan 8 '18 at 0:25
  • $\begingroup$ Okay, I understand now. That sounds more clear. Thanks! $\endgroup$ – numericalorange Jan 8 '18 at 1:20
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In your proof that $A\cup\varnothing=A$, we can just argue the following: for $x\in A\cup\varnothing$ we have $x\in A$ or $x\in\varnothing$. Of course, $x\notin\varnothing$ by definition of $\varnothing$, so we must have $x\in A$. Hence, $A\cup\varnothing\subseteq A$. For any sets $A$ and $B$ we clearly have $A\subseteq A\cup B$, and this is true when $B=\varnothing$. Therefore, $A\cup\varnothing=A$.

Your proof for $A\cap\varnothing=\varnothing$ looks good.

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  • $\begingroup$ Thanks! The first proof was a little hard for me, but thank you for making it clear. $\endgroup$ – numericalorange Jan 8 '18 at 1:21

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