# What is the meaning of "adding out" in Boolean algebra? I'm having a problem with what "adding out" means.

I understand that for the three first products in the second step, in the expression "(X + Y')*Z", "*Z" is replaced by adding the terms on the right side of the sum.

For the last three products, they add the terms on the right sight of the sum to "Z".

The problem is that I really don't understand why. Is "adding out" another procedure?

This is from the book Digital design: principles and practices, 4th edition, by Wakerly, page 202.

Adding out is done here to get the expression as a product of sums. Here is how it is obtained:

Using De Morgan's Law, we get:

$$\(X + Y')' = X'.Y\$$

Let $$\X + Y' = A. ---(1) \$$ Then $$\X'.Y = A'\$$

Substituting (1) in the expression for F, we get

$$\ F = A.Z + A'.Z' ---(2)\$$

Applying De Morgan's law, we get:

$$\ F = (A.Z)'.(A'.Z')'\$$

Applying De Morgan's law, we get:

$$\ F = (A' + Z').((A')' + (Z')') \$$

Because $$\(A')' = A\$$, we get

$$\ F = (A' + Z').(A + Z) \$$

$$\ F = A'.A + Z'.A + A'.Z + Z.Z' \$$

$$\ F = Z'.A + A'.Z \$$

Applying De Morgan's law, we get

$$\ F = ((Z')' + A).((A')' + Z') \$$

$$\ F = (A' + Z).(A + Z') \$$

Substituting $$\ (1) \$$ we get

$$\ F = (X'.Y + Z).(X + Y'+ Z') \$$

Since $$\ A.B + C = (A + C).(B + C) \$$ (Distributive Law),

We get:

$$\ F = (X' + Z).(Y + Z).(X + Y' + Z') \$$