# Discrete Math Proof: A cup B. The problem I'm having trouble proving is the following: A cup B=(A cap BC) cup (AC cap B) cup (A cap B) , where C denotes complement of a set

Discrete Math Proof: $A\cup B$
The problem I'm having trouble proving is the following:
$A\cup B=\left(A\cap BC\right)\cup \left(AC\cap B\right)\cup \left(A\cap B\right)$, where C denotes complement of a set
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umgangistbf
Step 1
Intuitively the result says that if an element is in $A\cup B$, then either the element is in A and not B, B and not A, or in both A and B.
Step 2
The arguement can then proceed as follows:

Ethen Frey
Step 1
First, you should not begin your proof by stating what you intend to prove. Instead of starting by writing $A\cup B=\left(A\cap {B}^{C}\right)\cup \left({A}^{C}\cap B\right)\cup \left(A\cap B\right)$ and aiming to reach $A\cup B=A\cup B$, you should start with $\left(A\cap {B}^{C}\right)\cup \left({A}^{C}\cap B\right)\cup \left(A\cap B\right)$ and show that it equals (through rewriting it in various forms that are the same set) as $A\cup B$.
(You can also show that two sets are equal by showing that each is a subset of the other; I don’t mean that my suggestion is the only approach, but it’s the appropriate way to write the proof using the particular mathematical argument you are presenting.)
This may seem like a silly question of style, but the fact is that you have only explained that if what you want to prove happens to be true, then it must follow that $A\cup B=A\cup B$. This is not a proof. It would also follows that $A\cup B=A\cup B$ even if what you began with is false. Any assumptions, even false ones, logically imply a true statement. A conjecture is not proven true by showing that you can use it to conclude a true statement. You have to show that you can proceed logically from true statements to prove it.
Step 2
Second, you should add some parentheses. At one point (after you use the distributive law), you write the expression $A\cap \left({B}^{C}\cup B\right)\cup \left({A}^{C}\cap B\right)$. You intend for this to mean $\left(A\cap \left({B}^{C}\cup B\right)\right)\cup \left({A}^{C}\cap B\right)$, and it’s safest to say so, because combinations of $\cup$ and $\cap$ may give different results depending on the order in which the operators are evaluated.