1. A\(B\C) = (A\B) U (A \C)
Can someone help me to prove this ? :/
Unfortunately, you have not given us any indication of what level of mathematics you have had exposure to. I recommend taking a stab at it, showing us your work and where you are stuck, and then we can help you. Are you familiar with set complements? What theorems have you already proven?
Sorry for the first post,
I am trying to do my discrete maths HW, and there is 2 exc left.
I need to show proofs from both sides.
First one:
Second:
And i can only use definitions and bullian algebra:
this is an example how we do things:
Ir - mean n, ARBA means u.
I dont have enough examples to prove this :/
No, that's wrong. There is a "d" in B so there is no "d" in A\B. A\B= {b, c}.
Yes, those are correct. And (A\B)U(A\C)= {b,c}U{b, c, d, f}= {b, c, d, f} which is NOT the same as A/(B\C).A\C = {b, c, d, f} B\C = {d, f}
A\(B\C) = {a, b, c, e}
Plato's point was that you can't prove this "theorem", it is NOT true!
You could prove
$$A\setminus (B \setminus C) = (A\setminus B) \cup (A \cap C)$$
$$\begin{align*}A\setminus (B\setminus C) & = A \setminus (B \cap C^c) \\ & = A \cap (B\cap C^c)^c \\ & = A\cap (B^c \cup C) \\ & = (A \cap B^c) \cup (A \cap C) \\ & = (A \setminus B) \cup (A \cap C)\end{align*}$$