I am given a function $f(W,X,Y,Z)$ that only outputs $1$ if exactly three inputs are $1$.
The AND-OR function is easy enough to find: $$ W'X\,YZ + WX'YZ + WX\,Y'Z + WX\,YZ' $$
I think that function is minimal as I can't find a way to simplify it I'm not supposed to be using K-maps yet, so I'm struggling with finding the OR-AND form. I've tried negating the AND-OR function as the complement of SoP is supposed to be PoS, which is the OR-AND function I think, but that's not the correct answer and I'm not sure why.
I've also tried writing out the entire formula in terms of zeroes instead of $1$s, i.e. $$ (W'+X'+Y'+Z')(W'+X'+Y'+Z)\cdots(W+X+Y+Z) $$ and I know I've done that correctly, but it gets too complicated for me to simplify quickly. I'm not sure where to go.
EDIT: I've been staring at the problem for a while so took a break and came back. I think I've got it worked out at this point though.
By first using uniting theorems ((X+Y)(X+Y')=X), I get it to a form where I can apply the second distributive law ((X+Y)(X+Z)=X+YZ) to cancel literals and then factor back out into a simpler form before repeating.
I got this as my answer, which matches what I would get from the K-Map:
(W+X)(W+Y)(W+Z)(X+Y)(X+Z)(Y+Z)(W'+X'+Y'+Z')
In my mind, the basic simplification for both AND-OR and OR-AND forms is, $$xy + xy' = x, \; (x+y)(x+y') = x$$
which can be interpreted as: if there are two input cases which differ in only one input but have the same output value, then they can be combined into a single case excluding that differing input. Generalizing for multiple differing inputs, if there is a set of input cases having the same output value and which run through all possible combinations of the differing inputs, then the set can be combined into a single case of the constant inputs. Alternatively, if you fix some inputs and the output is known regardless of the value of the other inputs, then you can write a single input case.
Based on the above, I don't see any further simplification to your AND-OR expression. For OR-AND, let's think of the complementary definition for $f(W,X,Y,Z)$: only outputs $0$ if zero, two, three, or four inputs are $0$. We can infer that if any two of the inputs are zero, the output is guaranteed to be $0$ regardless of the other inputs. So the two, three and four $0$'s cases can be represented by just the two $0$'s case, meaning we just have to write the zero and two $0$'s cases for the expression, $$(W' + X' + Y' + Z')(W + X + Y' + Z')(W + X' + Y + Z')(W + X' + Y' + Z)(W' + X + Y + Z')(W' + X + Y' + Z)(W' + X' + Y + Z)$$