Electrical – simplify A’B’C’+A’B’C+A’BC’+A’BC+ABC into minimal 1st canonical form

I question the real value of your question, since noone ever has only simple NOR gates to work with in practice. There are algorithms for finding near-optimal minimised solutions to less trivial expressions. https://en.m.wikipedia.org/wiki/Espresso_heuristic_logic_minimizer

Brute force is the best approach for a small problem, I fear.

I am going to work through this to the best of my ability - according to the excellent hint in the answer from Tom Carpenter's answer.

We start with

\$X = A_2 \overline A_1 + \overline Q_1 A_2 + Q_1 \overline A_2 A_1\$

which we can write as

\$ X = A_2(\overline A_1 + \overline Q_1) + Q_1 \overline A_2A_1 \$

and hence

\$ X = A_2(\overline{\overline{\overline A_1 + \overline Q_1}}) + Q_1 \overline A_2A_1 \$

then we can remove line redundancy and apply de Morgan's rule

\$ X = A_2(\overline{ A_1 Q_1}) + Q_1 \overline A_2A_1 \$

now add two more bars over the entire statement

\$ X = \overline { \overline { A_2(\overline{ A_1 Q_1}) + Q_1 \overline A_2A_1 }}\$

and then reapply de Morgan's rule

\$ X = \overline { \overline { A_2(\overline{ A_1 Q_1})} \cdot \overline {Q_1 \overline A_2A_1 }}\$

We can then draw this out as:

This is just 5 gates, and going through the truth table seems to show that the functionality is as required.