Help with simplifying boolean functions algebraically

Question

I have 2 boolean functions that I am having some difficulty solving algebraically.

NOTE: ~ means NOT, & means AND, + means OR

1) $(\sim b~\&~\sim d)+(b~\&~\sim c~\&~d)+(b~\&~c~\&~d)+(a~\&~\sim b~\&~d)$ Here is what I got:

$$\begin{align*} &=(b~\&~d)~\&~(\sim c+c)+(\sim b~\&~\sim d)+(a~\&~\sim b~\&~d)\\ &=(b~\&~d)+(\sim b~\&~\sim d)+(a~\&~\sim b~\&~d)\\ &=(a~\&~\sim b~\&~d) \end{align*}$$

How can this be? Doesn't the variable $c$ have to exist as well in the simplification for the function to be logically equivalent?

I have a similar issue with the next question:

2) $(a+b)~\&~(a+\sim b)$

Using the boolean distributive identity of $A + B \cdot C = (A + B) \cdot (A + C)$

$$\begin{align*} &=a+(b~\&~\sim b)\\ &=a \end{align*}$$

Once again, don't I need $b$ or $\sim b$ in the simplification?

Please help. Thank you!

Answer 1

It’s quite possible for a variable to disappear in a simplification. Just as in ordinary algebra you can have $ab-(a-1)(b-1)-b+1=ab-(ab-a-b+1)-b+1=a$, so in Boolean algebra you can have $(a+b)~\&~(a+\sim b)=a+(b~\&~\sim b)=a$.

However, your first calculation isn’t entirely correct: $(b~\&~d)+(\sim b~\&~\sim d)+(a~\&~\sim b~\&~d)$ is not equivalent to $(a~\&~\sim b~\&~d)$. The first expression is true when $b$ and $d$ are both true; the second is not.

Added: Here’s the truth table for $(a+b)~\&~(a+\sim b)$; I’ve used $1$ and $0$ for true and false, respectively.

$$\begin{array}{c|c|c|c|c} a&b&a+b&a+\sim b&(a+b)~\&~(a+\sim b)\\ \hline 0&0&0&1&0\\ 0&1&1&0&0\\ 1&0&1&1&1\\ 1&1&1&1&1 \end{array}$$

As you can see, the column for $(a+b)~\&~(a+\sim b)$ is identical to the column for $a$, so the two expressions must be equivalent.