Proving A - (B ∩ C) = (A - B) ∪ (A - C) in Discrete Math

[pyronova]

Show that A - (B intersection C) = (A - B) union (A - C)
I went about this completely around on a test but here is what I have

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

Easy problem but confused..thanks

 

[PeroK]

Show that A - (B intersection C) = (A - B) union (A - C)
I went about this completely around on a test but here is what I have

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

Easy problem but confused..thanks

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

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

 

[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.

 

[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}

 

[PeroK]

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}

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

That is the error.

 

[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 I am lost

 

[PeroK]

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 I am lost

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.