Electronic – Expression and gate not gate output

I think this is a solution, after playing with the expression a bit.

^{simulate this circuit – Schematic created using CircuitLab}

Derivation:

$$ AB\overline{C} + A\overline{B}C + \overline{A}BC + ABC$$ $$ = AB(C+\overline{C}) + A\overline{B}C + \overline{A}BC $$ $$ = AB + A\overline{B}C + \overline{A}BC $$ $$ = AB + C\cdot(A\overline{B} + \overline{A}B) $$ $$ = \overline{\overline{AB} \cdot \overline{C\cdot (A\overline{B} + \overline{A}B)}} $$ $$ = \overline{\overline{AB} \cdot \overline{C\cdot (A+B)\cdot\overline{AB}}} $$ Turning the 3-input NAND into 2 2-input NANDs and an inverter: $$ = \overline{\overline{AB} \cdot \overline{C\cdot \overline{\overline{(A+B)\cdot\overline{AB}}}}} $$

There may have been a wire in the wrong place or something. I did have it spread across three breadboards. I sketched everything out in a schematic, then pulled all the wires out and started over, following the schematic. Then things worked! Here's the schematic I used. The LED only lights up once the D input to the first flip-flop has gone HIGH and then LOW, which is how it should work.