I have the following formula: $(neg Aland Bland C)vee (neg Aland Bland neg C)vee (neg Aland neg B)vee (Aland C)vee (Alandneg C)$

After I did a Karnaugh Map for this formula I found out it is a tautology (in other words – all squares in the map are filled with ones). What is the minimal disjunctive normal form of this formula then?

I’ll show you how to prove the statement is tautology without Karnaugh’s map:

$(1)$ **distribution:**

$$(neg Aland Bland C)lor (neg Aland Bland neg C)equiv(neg Aland B)land(Clorneg C)equiv(neg Aland B)$$

$(2)$**distribution again:**

$$(neg Aland B)lor(neg Alandneg B)equivneg Aland(Blorneg B)equivneg A$$

$(3)$**distribution once again:**

$$(Aland C)lor(Alandneg C)equiv Aland(Clorneg C)equiv A$$

$$$$

$$underbrace{underbrace{(neg Aland Bland C)lor(neg Aland Bland neg C)}_{neg Aland B}lor(neg Aland neg B)}_{neg A}lorunderbrace{(Aland C)lor(Alandneg C)}_{A}equivneg Alor Aequiv 1$$