Proof: A'-B' = B-A (Double Inclusion) | Set Theory

[@tomkerruish2982]

You could also apply the definition of set difference, along with the commutativity of set intersection and the fact that (X')' = X for any set X (set complement is an involution). Explicitly:
A' - B' = A' int (B')' = A' int B = B int A' = B - A
However, your method of showing mutual set inclusion is more broadly applicable to such problems.
Good video!

[@inmathswetrust5293]

I would taught my students your method but if it was me solving it, I would prefer video's method

[@WrathofMath]

Thanks! I like your method a lot! Those string of equality type set proofs are always satisfying.

[@tomkerruish2982]

@@WrathofMath Thank you! I was likely inspired by Conway's 'one-line' proofs from On Numbers and Games, which are simply strings of equalities, often more than one line long.

[@tahiyaorsha2564]

Thanks for the video. I was confused so i searched and now I'm never going to make mistake. ❤️

[@marleny750]

i found this in a book
A minus B is equal to C times five.
can you explain it to me please?

[@Shaurya-Kushwaha]

Sir can I solve it like this ? : -
A' = U-A , B' = U-B
A'-B'= (U-A)-(U-B)
= U-A-U+B
=U-U+B-A = B-A

[@bhikaricharansahu4206]

Thankyou for explain this problem in easy method

[@WrathofMath]

Glad it helped, thanks for watching!

[@vinebadboi001]

Wow❤ thnks sir ........Like this video .Upload a lot of video ..Ok sir🎉

[@WrathofMath]

Thank you! Always more on the way!

[@coolATHARV0010]

Love from India sir 🙏🙏

[@kartikmishra8615]

Thanks

[@Arzoo-wc5uk]

Well explained...

[@WrathofMath]

Thank you!

[@vanshdeep-vy9ho]

Gora

[@quak1037]

THANKS

[@WrathofMath]

You're welcome!