Proof of sets A and B involving set theory, showing: $(B^c - AB)^c = B$ Use set algebra rules to show why the complement of $(B^c - AB)^c = B$

=> Let x be an object 

Assume $x\in (B^c -AB)^c $ or $x \notin (B^c - AB)$

So then $x \in B $ but $x\notin AB$, therefore $x \in B$

&lt;= Let x be an object

Assume $x \in B$ therefore $x \in B$ but  $x \notin AB$

In other words, $x \in (B^c)^c$ and $x \in (AB)^c$ 

or $x \in (B^c-AB)^c$

I feel like I'm on the right track but these can be cleaned up more
This looks completely fine to me. 

You're doing the definition of subset, each way ( implies-subset relation ) and showing the result follows.

You haven't written it in the most easy to follow form, but it is correct.

I prefer stuff like:


  Assume $x\in(\ldots)^C$  
  
  
  Then  $x\in\ldots$
  
  $\vdots$
  
  Thus $x\in(...)^C\implies x\in\ddots$ which by the implies-subset relation means $(\ldots)\subseteq\ddots$
  


But without the bullet points (cannot do indents here)
This looks completely fine to me. 

You're doing the definition of subset, each way ( implies-subset relation ) and showing the result follows.

You haven't written it in the most easy to follow form, but it is correct.

I prefer stuff like:


  Assume $x\in(\ldots)^C$  
  
  
  Then  $x\in\ldots$
  
  $\vdots$
  
  Thus $x\in(...)^C\implies x\in\ddots$ which by the implies-subset relation means $(\ldots)\subseteq\ddots$
  


But without the bullet points (cannot do indents here)