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Let A and B be subsets of a universal set U. If $A \subseteq B,$ then $B' \subseteq A'$

For the above question I'm trying to solve it by contrapositive proof. I was suggested by some friends that it can also be solved using proof by contradiction. Can someone tell me how this could be done by proof by contradiction?

Chappers
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User76611
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  • Prove the statement directly. Take an element in $B’$ and prove that it must be in $A’$. – Clayton Oct 04 '20 at 19:27
  • In a proof by contradiction you take the hypothesis and assume the negation of the result is true, then arrive to an absurdum, usually the negation of the hypothesis (the contradiction). Since we do not want any statement to be true and false at the same time, then the negation of the result cannot be assumed true, so it must be false and the result must be true (this is using the law of excluded middle: any statement is either true or false; there is no other option). Proof by contrapositive and contradiction are equivalent when the Law of Excluded Middle is assumed. – Darsen Oct 04 '20 at 19:36

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