Process of simplifying boolean expression 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?
$$\overline{ab}cd + a\overline{b}cd = \overline{b}cd$$
 A: As you seem to have suspected, there’s more than one simplified form. Here’s an algebraic derivation of a different one:
$$\begin{align*}
\color{purple}{\overline{ab}c\overline{d}}&+\color{purple}{\overline{ab}cd}+\overline{a}b\overline{c}d+\overline{a}bcd+a\overline{b}cd+ab\overline{c}d\\
&=\color{purple}{\overline{ab}c(\overline{d}+d)}+\overline{a}b\overline{c}d+\overline{a}bcd+a\overline{b}cd+ab\overline{c}d\\
&=\color{purple}{\overline{ab}c}+\color{green}{\overline{a}b\overline{c}d+\overline{a}bcd}+a\overline{b}cd+ab\overline{c}d\\
&=\overline{ab}c+\color{green}{\overline{a}b(\overline{c}+c)d}+a\overline{b}cd+ab\overline{c}d\\
&=\overline{ab}c+\color{green}{\overline{a}bd}+a\overline{b}cd+ab\overline{c}d\\
\end{align*}$$
A: 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
$$\overline{b}cd+b\overline{c}d+\overline{a}bd+\overline{a}\overline{b}c$$
