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Prove:

$1)$ $A\cup B=B\cup A$

$2)$$ A\cup A=A$

$3)$ $(A\cup B)^{c}= A^{c}\cap B^{c}$

I'm not sure whether my proofs are fine:

$1)$ Let's suppose the opposite, i.e. $A\cup B\neq B\cup A$

$a)$ $A\cup B\nsubseteq B\cup A \iff x\in (A\cup B) \Rightarrow x\notin (B\cup A) \iff (x\in A\lor x\in B) \Rightarrow (x\notin B \land x\notin A) \Rightarrow\Leftarrow$

$b)$ $B\cup A\nsubseteq A\cup B \iff x\in (B\cup A)\Rightarrow x\notin(A\cup B) \iff (x\in B \lor x\in A) \Rightarrow (x\notin A \land x\notin B) \Rightarrow\Leftarrow$

$2)$ Let's suppose the opposite, i.e. $A\cup A\neq A$

$a)$ $x\in (A\cup A)\Rightarrow x\notin A \iff (x\in A\lor x\in A)\Rightarrow x\notin A \iff (x\in A)\Rightarrow (x\notin A) \Rightarrow\Leftarrow$

$b)$ $x\in A \Rightarrow x\notin(A\cup A) \iff(x\in A) \Rightarrow (x\notin A \land x\notin A) \iff x\in A \Rightarrow x\notin A \Rightarrow\Leftarrow$

$3)$ Let'suppose the opposite, i.e. $(A\cup B)^{c}\neq A^{c}\cap B^{c}$

$a)$ $x\in (A\cup B)^{c} \Rightarrow x\notin (A^{c}\cap B^{c}) \iff x\notin (A\cup B)\Rightarrow (x\notin A^{c} \lor x\notin B^{c}) \iff (x\notin A \land x\notin B) \Rightarrow (x\in A \lor x\in B) \Rightarrow\Leftarrow$

$b)$ $x\in(A^{c}\cap B^{c})\Rightarrow x\notin(A\cup B)^{c} \iff (x\in A^{c} \land x\in B^{c}) \Rightarrow x\in (A\cup B) \iff (x\notin A \land x\notin B) \Rightarrow (x\in A) \lor (x\in B) \Rightarrow\Leftarrow$

lmc
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1 Answers1

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I think your proofs are fine, although you don't have to assume the opposite and go for a contradiction. For example, to prove $A \cup B \subset B\cup A$, you could say:

Take $x \in A\cup B$. Then $x\in A\lor x\in B$. So $x\in B\lor x\in A$ (because $\lor$ is commutative), so $x \in B\cup A$.

As a general rule, you should prefer a direct proof over a proof by contradiction.

Mike Pierce
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