# Discrete Math Question

1. Oct 23, 2014

### pyronova

• Member warned about not using the homework template
Show that A - (B intersection C) = (A - B) union (A - C)

Right Hand Side = ( A - B) union ( A - C)
= (A intersection B) union (A intersection C)
= A - (B intersection C) ?

Easy problem but confused..thanks

2. Oct 23, 2014

### PeroK

Note that $A - B = A \cap B'$

You could solve this by showing that $x \in LHS \ \ \Rightarrow \ \ x \in RHS$ and vice versa.

3. Oct 23, 2014

### GFauxPas

Not sure why this is in the Engineering forum.
Mod note: It's now in the Precalc section.
One way to prove equality of sets is to show each is a subset of the other. So you want to prove that:
$a \land \neg (b \land c) \implies (a \land \neg b) \lor (a \land \neg c)$
$(a \land \neg b) \lor (a \land \neg c) \implies a \land \neg (b \land c)$
I'd use DeMorgan's on the LHS on the first and on the RHS on the second, personally.
edit: or factor, if your professor allows you to do it that way.

Last edited by a moderator: Oct 23, 2014
4. Oct 23, 2014

### Gilbert

I think there is a mistake in your equation : (A-B)∪(A-C) ≠ (A∩B)∪(A∩C).
Here it's a general example to verify it:
A={a,b,c,d} , B={a,c} and C={c,d}

A-(B∩C) = {a,b,c,d} - ({a,c}∩{c,d}} = {a,b,c,d} - {c} = {a,b,d}

And (A-B)∪(A-C) = {b,d}∪{a,b} = {a,b,d}

Then : A-(B∩C) = (A-B)∪(A-C).

But (A∩B)∪(A∩C) = {a,c,d} ≠ {a,b,d}

5. Oct 23, 2014

### PeroK

$A - B = A \cap B',$ $A - B \ne A \cap B$

That is the error.

6. Oct 24, 2014

### pyronova

So I have..
x∈RHS <---> x∈(A-B)U(A-C)
<----> x∈ A ^ xnot∈ B U x∈ A ^ xnot∈ C
<---->x∈ A∩B' U A∩C'
<----> x∈ A - (B' ∩ C')
<----> x∈ A - (B ∩ C) ?
man im lost

7. Oct 24, 2014

### PeroK

Stick with it. You are making progress. I would try it first like this

$x \in (A-B) \cup (A-C)$ iff (x is in A and not in B) or (x is in A and not in C)

Iff (x is in A) and (x is not in B or not in C) (check this step then keep going)

In this problem you really have to think about what the symbols mean. Rather than just trying to manipulate them according to some rules.