Presented with the Boolean function $F$, I must reduce this to its most simple form. I know the answer is $(A+BD)$, but I am lost on how to get there.
$F=((((A'B)'+D')'+A)(((A'B)'+D')'+C'))+((D'+D)(C'C)) + AC$
Boolean Algebra – How to Reduce Boolean Algebra Expressions
boolean-algebralogic
Related Solutions
For $(1)$, we can use two forms of the distributive law - one you used, as well as: $$X + YZ = (X + Y)(X+Z)$$
$$\begin{align} AC + A'B'C = C(A + A'B') & = C\Big((A+A')(A+B')\Big) \\ \\ & = C\Big((1)(A+B')\Big) \\ \\ & = C(A+B') \\ \\ &= AC + B'C\end{align}$$
For $(2)$ we can use the identity given by: $X = X(1) = X(\underbrace{Y + Y'}_{\large =\, 1}) = XY + XY'\tag{ID}$:
$$\begin{align} A'B'+A'BC'+(A+C')' & = A'B'+A'BC'+A'C \\ \\ & = A'(B'+BC'+C)\\ \\ & = A'\Big((B'C + B'C') + BC' + (BC + B'C)\Big) \tag{ID}\\ \\ & = A'\Big(\underbrace{BC + B'C + BC' +B'C'}_{\large = 1}\Big)\\ \\ & = A' \end{align}$$
In the second to last line, note that one of $BC, B'C, BC', B'C'$ is necessarily true (since there are four possible truth value assignments to two variables).
Best Answer
If you use these, the expression reduces to,
$A'BD+AC'+AC = A'BD+A(C+C') = A+A'BD$
Now,
The expression thus becomes, $A+A'BD = A+BD$