In your post, you say:
2. (AB)'=A'+B'
3. (A+B)'=A'B'
I think it should be (A+B+C)'. <== I don't know where this fits in
then you provide four combinations:
A'+B'+C' and (A+B+C)' or A'B'C' and (ABC)'
Looking at each one,
A'+B'+C' = (ABC)' (flipping 2 around and extending to three terms)
(A+B+C)' = A'B'C' (from 3, extending to three terms)
A'B'C' = (A+B+C)' (flipping 3 around and extending to three terms)
(ABC)' = A'+B'+C' (from 2, extending to three terms)
so the first and last are the same, and the middle two are the same.
If we allow exor gates then here is a solution.
1st col 4th map: (A exor C) + AB
2nd col 2nd map: (A exor B) + ABC
2nd col 3rd map: !(A exor C) + AB, alternatively (A exnor C) + AB
Edit:
Take col 1 map 4 for example. The first two columns have a pattern I recognize as an XOR gate. First row is 01, second row is 10. Now I look at the boxes with '1', A changes when jumping between the two. Look for another variable that is also changing; in this case it is C. A and C are either 01 or 10 in the '1' boxes: that is the characteristics of an XOR gate. Only two '1' remain in the map and they are grouped into the term AB.
Now col 2 map 2: Here are two groups that will work for an XOR gate; col 1 & 4, and col 3 & 4. The first group produces an XOR gate, the second an XNOR gate. In the XNOR gate, both inputs must be equal to produce a '1' on the output.
Finally col 2 map 3. This map looks like col 1 map 4, except the XOR pattern is reversed. That means we use an XNOR gate instead of the XOR in the first map, or we add an inverter to the XOR output.
By the way, regarding your equation for col 1 map 4, notice that the two bottom corners both have '1'. You can group them to produce the term A!C, reducing your third term to two variables.
Best Answer
Wolfram Alpha appears to give the right answer. I used their Boolean Algebra widget since for some reason it didn't recognize the same formula entered on their main page as a Boolean algebra expression.
I entered:
(A and B and C) or (A and B and D) or (A and C and D) or (not B and not C and D) or (A and C and not D) or (not A and not B and D)
and it returned:
Interesting that they ask you to enter equations using AND, OR, NOT, XOR, NAND, NOR etc. but the results are displayed as shown above. Anyhow the first result (DNF, which stands for disjunctive normal form) matches your result.