Talk:Union with Complement

This proof, Union with Complement, depends on Union with Relative Complement.

Union with Relative Complement depends on Set Difference Union Second Set is Union.

Set Difference Union Second Set is Union depends on Union with Complement.

That's a problem, I think? --GFauxPas 10:56, 7 December 2011 (CST)

You're right. I wrote a new proof for Union with Relative Complement. Can you check if it works now? --Dan232 12:29, 7 December 2011 (CST)

I think it works, but I would add a few steps to justify it more. I tried to redo it to fix some things, but I ran out of time, I have to do something else, sorry!. I would use:

$\complement_S \left({T}\right) \cup T\subseteq S$

then turn it into a universal statement by definition of subset, then use Universal Instantiation to turn it into a conditional. Then use Proof by Cases and go back to subset with Universal Generalisation. Do the same the other way, and then use definition of set equality. As it is set up now, the proof doesn't work for all $x$, only for the one you picked. But maybe I'm being too Bourbakian. --GFauxPas 13:38, 7 December 2011 (CST)

No such thing as being too Bourbakian. Nobody here is Bourbakian enough. :-)

Anyway, I have prime.moverified the proof and see how you find it now. Good work noticing (and fixing) that embarrassing circularity. --prime mover 13:58, 7 December 2011 (CST)

How 'bout now? --GFauxPas 16:27, 7 December 2011 (CST)

Okay I think it's done now, thank you very much Dan 232 and prime.mover. Should we leave Proof 1 up? --GFauxPas 17:12, 7 December 2011 (CST)

No reason not to. --prime mover 00:32, 8 December 2011 (CST)

Changed my mind. Occurred to me that this is a fundamental result. --prime mover 01:20, 8 December 2011 (CST)