# Simplification in a Boolean algebra

I tried to do question 7 of chapter 5-2 in Boolean Algebra and its Applications, by J. Eldon Whitesitt. Starting from table 5-1 and calling the function represented by the table $f$, I get

$$f = AB'C + AB'C' + ABC' + A'B'C.$$

From the solutions at the end of the book, this should simplify to

$$f = AC' + B'C,$$

which makes sense when I draw the Venn diagrams representing the sets equivalent to the formula. However, when I proceed algebraically, I am only able to simplify to

$$f = AB' + AC' + B'C$$

or

$$f = (A'B + A'C' + BC)'.$$

How can I make the $AB'$ term vanish in my first simplified form?

The Consensus Theorem (see Wikipedia entry) does this in 1 step:

Consensus Theorem

$XY+Y'Z+XZ \Leftrightarrow XY+Y'Z$

We can derive the Consensus Theorem (and thus your equivalence as well) from some more basic principles:

$XY+Y'Z+XZ= \text{ Adjacency}$ (Adjacency says $PQ+PQ'=P$)

$XY+Y'Z+XYZ+XY'Z= \text{ Absorption}$ ($XY$ absorbs $XYZ$ and $Y'Z$ absorbs $XY'Z$)

$XY+Y'Z$