Process of simplifying boolean expression

Question

I have an expression:

$$y = \overline{ab}c\overline{d} + \overline{ab}cd + \overline{a}b\overline{c}d + \overline{a}bcd + a\overline{b}cd + ab\overline{c}d$$

I've constructed the circuit for this expression and found a couple of redundancies. I've deduced a/the simplified form of the expression is:

$$y = \overline{ab}c\overline{d} + \overline{b}cd + b\overline{c}d + \overline{a}bcd$$

It's for a four-bit prime checker. For reference, $a$ corresponds to $2^3$. I've deduced this is a/the simplified form because a couple of duos of prime numbers don't care for $a$ ($0011_2$ and $1011_2$, for example). The output is true irrespective if $a$ is true in these cases. Because both $\overline{ab}cd$ and $a\overline{b}cd$ are true, the $a$ is redundant.

How can I deduce it algebraically instead of investigating the circuit? Wild guess, but is the below true?

$$\overline{ab}cd + a\overline{b}cd = \overline{b}cd$$

Answer 1

As you seem to have suspected, there’s more than one simplified form. Here’s an algebraic derivation of a different one:

$$\begin{align*} \color{purple}{\overline{ab}c\overline{d}}&+\color{purple}{\overline{ab}cd}+\overline{a}b\overline{c}d+\overline{a}bcd+a\overline{b}cd+ab\overline{c}d\\ &=\color{purple}{\overline{ab}c(\overline{d}+d)}+\overline{a}b\overline{c}d+\overline{a}bcd+a\overline{b}cd+ab\overline{c}d\\ &=\color{purple}{\overline{ab}c}+\color{green}{\overline{a}b\overline{c}d+\overline{a}bcd}+a\overline{b}cd+ab\overline{c}d\\ &=\overline{ab}c+\color{green}{\overline{a}b(\overline{c}+c)d}+a\overline{b}cd+ab\overline{c}d\\ &=\overline{ab}c+\color{green}{\overline{a}bd}+a\overline{b}cd+ab\overline{c}d\\ \end{align*}$$

Answer 2

While not strictly algebra a Karnaugh map is an good way to arrival at a minimal expression. Algebra alone will not allow you to determine if there are not smaller expressions.

| a=1 | a=0 | ---+---+---+---+---+--- | 0 | 0 | 0 | 1 | d=0 c=1 +---+---+---+---+--- | 1 | 0 | 1 | 1 | ---+---+---+---+---+ d=1 | 0 | 1 | 1 | 0 | c=0 +---+---+---+---+--- | 0 | 0 | 0 | 0 | d=0 ---+---+---+---+---+--- |b=0| b=1 |b=0| (Sorry, new to SE so not sure how to do a table properly).

For this problem there are 2 equivalent expressions using the least number of expressions.

$$\overline{b}cd+b\overline{c}d+\overline{a}cd+\overline{a}\overline{b}c$$ or $$\overline{b}cd+b\overline{c}d+\overline{a}bd+\overline{a}\overline{b}c$$