I have a boolean logic statement:
I get the truth table to derive the statement:
derived statement:
(-A^-B^C) V (-A^B^-C) V (-A^B^C) V (A^-B^C) V (A^B^C)
But when I apply de Morgan's to simplify it, I end up with nothing:
-(AVBV-C) V -(AV-BVC) V -(AV-BV-C) V -(AVBV-C) V -(-AV-BV-C)
=== -A V -B V C V -A V B V -C V -A V B V C V -A V -B V C V A V B V C
=== -A V -B V C V B V -C V A
=== 0
Is there another simplification that makes more sense?
From your truth-table it should be clear that you expression should not equal $0$ ... and by the way, your expression right before $0$ equals $1$, not $0$ .... but the truth-table also makes it clear it is not $1$ either. So, something clearly went wrong ...
Here is your mistake. You went from
$\neg (A \lor B \lor \neg C) \lor ....$
to
$\neg A \lor \neg B \lor C \lor ....$
You applied DeMorgan correctly from line 1 to 2, but did not apply it correctly from line 2 to 3 ... indeed, you'd just get back to line 1 if you were to apply it correctly.
Now, as far as simplifying your statement goes ... note that in every case that $C=1$, the expression evaluates to $1$ as well. So, all those rows in the truth table can be captured by just $C$. The only other case is the $\neg A \land B \land \neg C$, so you can immediately simplify the statement to
$C \lor (\neg A \land B \land \neg C)$
which further simplifies to
$C \lor (\neg A \land B)$
Looking at your truth table, it is pretty easy to see the expression
$$C \lor(\lnot A \land B)$$
works. The value of $C$ is identical to your output value for all but one line. The expression in parentheses accounts for that one exceptional line (and the line above it too).
