Bogus set theory proof

Question

I'm having trouble figuring out where I went wrong in this proof. I think it's to do with my understanding of things like $\cup$ and $\cap$ in that I don't really have a solid understanding of what they mean other than the venn diagram explanation.

I will "show" that $A-B\cup C=A-B\cap C$.

If $x\in A-B\cup C$ then $x\in A$ and $\not\in B\cup C$. Then $x\in A$ and $\not\in$ at least one of $B,C$. So $x\in A$ and $\in$ both $B,C$ except one. So $x\in A$ and $\not\in$ both $B$ and $C$. So $x\in A$ and $\not\in B\cap C$. So finally $x\in A-B\cap C$. Similarly I can run backwards to get $x\in A-B\cap C\implies x\in A-B\cup C$ so that $A-B\cup C=A-B\cap C$, which is clearly false.

Edit: I will write the part starting "similarly I can run backwards" in full to avoid any ambiguity. (Thanks to nombre for pointing this out.) Also everywhere in this post please read $A-B?C$ as $A-(B?C)$ (for $?\in \{\cup,\cap\}$).

So here is the full ending showing the other direction:

If $x\in A-B\cap C$ then $x\in A$ and $\not\in $ both $B$ and $C$. So $x\in A$ and $\in$ both $B$ and $C$ except for one. So $x\in A$ and $\not\in$ at least one of $B,C$. So $x\in A-B\cup C$.

Thanks

Answer 1

If something is not in $B \cup C$, then it is in neither $B$ nor $C$. Because if it was in $B$, then it is in $B$ or $C$, which is $B \cup C$. So the first mistake is in the second sentence.

You could show that $A - (B \cup C) \subseteq A - (B \cap C)$. Because the former one is what is in $A$, but not in $B$ nor $C$. The latter is what is in $A$, but not in at least one of $B$ and $C$.

I would encourage drawing a Venn-Diagram when doing such things. That makes it so much easier.

Answer 2

"And not in at least one of B,C. So $ x \in A$ and $x \in $ both A and B except one." That's the heart of the problem. $x\notin (B \cup C) \implies x \notin B $ and $ x \notin C$. x is in neither B nor C.

Do you see where you went wrong now?

Answer 3

You can't "similarly run backwards". If you try to detail that part, you'll see there is a problem. If $x\notin B\cap C$, it does not mean that $x\notin B\cup C$. For exemple if $x\in B$ but not $C$, you have $x\notin B\cap C$ but you don't have $x\notin B\cup C$.