1. ## Discrete maths proofs

1. A\(B\C) = (A\B) U (A \C)
Can someone help me to prove this ? :/

2. ## Re: Discrete maths proofs

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?

3. ## Re: Discrete maths proofs

Originally Posted by juliusxxc
1. A\(B\C) = (A\B) U (A \C)
Can someone help me to prove this ? :/
@juliusxxc, I absolutely agree with others that you must show some effort.

Suppose that $A=\{a,b,c,d,e,f\}~,~B=\{a,d,e,f\}~,\&~C=\{a,e\}$

Now you show us:
$A\setminus B=?$ , $A\setminus C=?$ , & $B\setminus C=?$

So $A\setminus (B\setminus C)=?$

4. ## Re: Discrete maths proofs

A\B = {b, c, d} A\C = {b, c, d, f} B\C = {d, f}
A\(B\C) = {a, b, c, e}

5. ## Re: Discrete maths proofs

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 :/

6. ## Re: Discrete maths proofs

Originally Posted by juliusxxc
A\B = {b, c, d}
No, that's wrong. There is a "d" in B so there is no "d" in A\B. A\B= {b, c}.

A\C = {b, c, d, f} B\C = {d, f}
A\(B\C) = {a, b, c, e}
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).

Plato's point was that you can't prove this "theorem", it is NOT true!

7. ## Re: Discrete maths proofs

well i still need to show it in the way that i showed you bellow.
Thanks for taking time and replying

8. ## Re: Discrete maths proofs

Originally Posted by juliusxxc
well i still need to show it in the way that i showed you bellow.
You are wasting your time as well as our time.
You cannot prove something true if it is false.
The example I posted shows that $A\setminus (B\setminus C)\ne (A\setminus B)\cup(A\setminus C)$ in general.

9. ## Re: Discrete maths proofs

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*}

10. ## Re: Discrete maths proofs

Originally Posted by juliusxxc
Could you explain what does the little c means ?
That's the complement of a set

$A^c = \{x : x \not \in A\}$