Simplify the boolean function $$Z=A\bar B \bar{C_i} + \bar A B \bar{C_i} + \bar A\bar B {C_i} + A B {C_i}$$

Question

I want to simplify the following boolean function:

$$Z=A\bar B \bar{C_i} + \bar A B \bar{C_i} + \bar A\bar B {C_i} + A B {C_i}$$

Here's my attempt:

\begin{align} Z &= A\bar B \bar{C_i} + \bar A B \bar{C_i} + \bar A\bar B {C_i} + A B {C_i} \\ & = \bar{C_i}(A \bar B + \bar A B) + C_i(\bar A \bar B + AB) \\ & = \bar C_i(A \oplus B) + C_i(A \equiv B) \end{align}

I thought this was the end of it but in my textbook it continues and has: \begin{align} Z &= A\bar B \bar{C_i} + \bar A B \bar{C_i} + \bar A\bar B {C_i} + A B {C_i} \\ & = \bar{C_i}(A \bar B + \bar A B) + C_i(\bar A \bar B + AB) \\ & = \bar C_i(A \oplus B) + C_i(A \equiv B) \\ & = A \oplus B \oplus C_i \\ & = A \equiv B \equiv C_i \end{align}

I'm confused about what happened between the third and fourth step. What boolean algebra rules are being used here?

Answer 1

When in doubt about booleans, just build a truth table.

Truth tables for XOR (\$\oplus\$):

   | 0 | 1
---+---+---
 0 | 0 | 1
---+---+---
 1 | 1 | 0

For "is equal to" (\$\equiv\$):

   | 0 | 1
---+---+---
 0 | 1 | 0
---+---+---
 1 | 0 | 1

As you can see \$A \equiv B\$ gives just the opposite result of \$A \oplus B\$ (the result is 1 for the first when it is 0 for the second, and vice-versa). This means that:

$$A \equiv B = \overline{A \oplus B}$$

You used several times the identity $$X\overline{Y} + \overline{X}Y = X \oplus Y$$

This means: If (X is true AND Y is false) OR (if X is false and Y is true) is the same as either X or Y is true, but not both, which is quite straightforward.

So now you get to this equation:

$$\overline{C_i}(A \oplus B) + C_i(A \equiv B) \\$$

Since \$A \equiv B\$ can be written as \$\overline{A \oplus B}\$, you can rewrite it to:

$$\overline{C_i}(A \oplus B) + C_i(\overline{A \oplus B}) \\$$

Which is a form of \$X\overline{Y} + \overline{X}Y\$, with \$X = C_i\$ and \$Y = A \oplus B\$.

So it can then be rewritten to:

$$C_i \oplus (A \oplus B)$$

As all these boolean operators are commutative, this be be rewritten as:

$$A \oplus B \oplus C_i$$

Answer 2

Observe that
\begin{align}\overline{\overline{A} \cdot\overline{B} + A\cdot B} = \overline{(\overline{A}\cdot\overline{B})}\cdot\overline{(A \cdot B)} = (A + B)\cdot(\overline{A}+\overline{B}) = A \cdot\overline{B} + \overline{A}\cdot B \end{align}

Hence

\begin{align} Z&=\overline{C_i} (A \oplus B) + C_i (A\equiv B)\\ &=\overline{C_i} (A \oplus B) + C_i\overline{(A \oplus B)} \\ &= C_i \oplus (A\oplus B)\\ &=A\oplus B\oplus C_i \end{align}

Answer 3

First, plus is an unusual operator for bool equations Wikipedia. I assumed you referencing a OR

With this assumtion I came to the result: no further reduction possible. For This I used a Karnaugh map

https://en.wikipedia.org/wiki/Karnaugh_map