I have the following formula: $(\neg A\land B\land C)\vee (\neg A\land B\land \neg C)\vee (\neg A\land \neg B)\vee (A\land C)\vee (A\land\neg C)$
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: $$(\neg A\land B\land C)\lor (\neg A\land B\land \neg C)\equiv(\neg A\land B)\land(C\lor\neg C)\equiv(\neg A\land B)$$ $(2)$distribution again: $$(\neg A\land B)\lor(\neg A\land\neg B)\equiv\neg A\land(B\lor\neg B)\equiv\neg A$$ $(3)$distribution once again: $$(A\land C)\lor(A\land\neg C)\equiv A\land(C\lor\neg C)\equiv A$$ $$$$ $$\underbrace{\underbrace{(\neg A\land B\land C)\lor(\neg A\land B\land \neg C)}_{\neg A\land B}\lor(\neg A\land \neg B)}_{\neg A}\lor\underbrace{(A\land C)\lor(A\land\neg C)}_{A}\equiv\neg A\lor A\equiv 1$$