# Basic boolean logic

#### fran1942

Joined Jul 26, 2010
58
Hello, I am trying to grasp this topic.
I have the following expression (see attached image).
I need to
i) simplify using deMorgans theorem
ii) then show how the expression could be implemented using only 2 input NOR gates.

I "think" I have the deMorgans simplification correct, but I am not sure how to do part ii.
Any help would be much appreciated.

#### lunapt

Joined Feb 2, 2013
12
write the equation so it has two OR gates, then negate the entire equation

#### fran1942

Joined Jul 26, 2010
58
thanks, but that is the problem.
I am not sure how to rewrite my deMorgan simplification as two NOR gates.

#### lunapt

Joined Feb 2, 2013
12
double negate the entire equation so you can deMorgans again to switch the gates

#### DerStrom8

Joined Feb 20, 2011
2,390
You used DeMorgan's theorem incorrectly. When you use DeMorgan's theorem, you split up the "not" bar and change the operation. For example:

(A+B)' = A'*B'

The ' symbolizes a "Not" bar. As you can see, the entire first part (A+B) is inverted, so you have a bar above all of it. You split the bar so that each part has the bar over it, and then you change the operation from OR to AND (+ to *).

EDIT: Sorry, you are correct. I misread the expression #### WBahn

Joined Mar 31, 2012
25,899
You actually only need a single 2-input NOR gate.

This is what you started with

J = [(G'*H')' + (H*I')]'

Applying DeMorgan's gets you

J = (G'*H')*(H*I')'

Applying DeMorgan's to the second term gets you

J = (G'*H')*(H'+I)

But at this point you just removed the parens around the second factor. You have to distribute one over the other.

Now take a step back and look at your original expression

J = [(G'*H')' + (H*I')]'

This is ALREADY a NOR gate!

J = NOR[(G'*H')', (H*I')]

So focus on the guts of this expression and in three easy steps it will reduce to the OR of just two terms.

#### DerStrom8

Joined Feb 20, 2011
2,390
You actually only need a single 2-input NOR gate.

This is what you started with

J = [(G'*H')' + (H*I')]'
That is not correct. The bar goes over the entire G*H expression. It is NOT G'*H', it is (G*H)'.

The OP's simplification is done correctly.

#### WBahn

Joined Mar 31, 2012
25,899
Oops. That's what happens when a nearly blind man that sees double and triple images has to switch between windows and remember what lines he saw where.

It's a shame, too, because my incorrect version simplifies so beautifully!

But the OP's simplification is still wrong. The mistake I made does not affect the need to distribute one factor over the other.

So hopefully this is the correct starting point.

J = [(G*H)' + (H*I')]'

This is still a NOR gate at the outermost level.

Applying DeMorgan's

J = (G*H)*(H*I')'

Applying DeMorgan's to the second term

J = (G*H)*(H'+I)

The OP has this

J = G*H*H'+I

Which is wrong, since AND takes precedence over OR and this would simplify to J=I

J = (G*H)*(H'+I)

J = (G*H)*H'+(G*H)*I

J = G*H*I

#### DerStrom8

Joined Feb 20, 2011
2,390
Applying DeMorgan's to the second term

J = (G*H)*(H'+I)

The OP has this

J = G*H*H'+I
Oops, you're right. Good catch!

#### WBahn

Joined Mar 31, 2012
25,899
A quick count, assuming I am correct that J=GHI, is that it will take six NOR2 gates.

#### WBahn

Joined Mar 31, 2012
25,899
Oops, you're right. Good catch!
Bwtween us we'll get it right! • DerStrom8