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There are several solutions to the proof $P(A\cup B)=P(A)\cup P(B)\rightarrow A\subseteq B\vee A\supseteq B$ already, however I would like to see if mine is correct:

Proof through contradiction:

Let us assume $A\nsubseteq B\vee A\nsupseteq B$ such that $a\in A\wedge a\notin B, b\in B\wedge b\notin A$.
However, this is impossible as $a,b\in P(A\cup B)=a,b\in P(A)\vee a,b\in P(B)$ and therefore, $P(A\cup B)=P(A)\cup P(B)\rightarrow A\subseteq B\vee A\supseteq B \blacksquare$

Does this proof work? If not, Why?

Henno Brandsma
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CluelessButCurious
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  • Don't you mean $P(A \cup B)=P(A)+P(B) $? The expression on the left represents a a number, while you are unioning numbers on the right, which does not have a clear definition AFAIK. – gary Jul 25 '17 at 18:58
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    I believe he means $\cup$. $P$ means, at least in my experience, the power set (and not just the cardinality thereof). – Alex Meiburg Jul 25 '17 at 18:59
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    I'm more accustomed to seeing $\mathcal{P}$ for power set, and regular $P$ for probability. – G Tony Jacobs Jul 25 '17 at 19:06
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    This was asked several dozen times by now. You're asking for a proof verification, so this is not entirely the same. But you can find proofs and compare yours to them all over the place. – Asaf Karagila Jul 25 '17 at 20:36
  • If you are doing a proof by contradiction, you should have an and symbol near the beginning. – Jacob Wakem Jul 26 '17 at 17:48

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The first line is OK. Then you write $a,b \in P(A \cup B)$ which is not true, only $a,b \in A \cup B$ does hold. Try $\{a,b\} \in P(A \cup B)$, which is true. Then $\{a,b\} \in P(A)$ leads to $\{a,b\} \subseteq A$ which implies $b \in A$, contradiction. (similarly when $\{a,b\} \in P(B)$). You did not specify where the contradiction comes from, but jump straight to the conclusion.

Henno Brandsma
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