# PoS-SoP convertion of function bellow (NOT using K-map)

Function F as: $$F=\overline{C} \space \space \overline{D} +A\overline{D} + A\overline{B}$$ is a SoP (sum of products) which is given by grouping K-map's 1's.

I tried to complement the function $$\F\$$ to get PoS (product of sums), like: $$\overline{F}= (C+ D) \cdot (\overline{A}+D) \cdot (\overline{A}+B)$$ but I checked the K-map's 0's to see if it's a correct PoS, but it is not even a PoS because it contains max terms that the correct PoS shouldn't have.

Does any one know where I'm wrong?

• Multiply it out, reduce and invert a second time. You've done DeMorgan's once. You have to do it twice to go from PoS to SOP. – StainlessSteelRat Nov 16 '18 at 17:31
• @StainlessSteelRat does it work? I did and it didn't work – parvin Nov 16 '18 at 17:32
• It works. I did it. Answer was the same as K-map. – StainlessSteelRat Nov 16 '18 at 17:32
• @StainlessSteelRat the answer is (C'+A')*(D'+A)*(B'+D') and when I draw the K-map it has wrong max terms. – parvin Nov 16 '18 at 17:34
• Correction: I get $f= (\overline B + \overline{D}) (A + \overline C) (A + \overline D )$ – StainlessSteelRat Nov 16 '18 at 18:17

$$F=\overline{C} \space \overline{D} +A\overline{D} + A\overline{B}$$

DeMorgan's

$$\overline F=\overline{\overline{C} \space \overline{D} +A\overline{D} + A\overline{B}}$$

$$\overline F= (C + D) (\overline A + D) (\overline A + B)$$

First two terms: $$\overline F= (C\overline A + D\overline A + CD + DD) (\overline A + B)$$

Simplify: $$\overline F= (C\overline A + D) (\overline A + B)$$

Finish multiplying out last term, reduce and take DeMorgan's.