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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?

PinkyWay
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TKN
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  • I'm sorry but I am a bit lost. Could you please just tell me what is the minimal disjunctive normal form of this tautological formula? I would deduce the solution from that. – TKN Jan 28 '20 at 14:31
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    Since you know it's a tautology, $\top$ is the minimal disjunctive normal form, also the minimal conjunctive normal form. – Ethan Jan 29 '20 at 02:43

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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$$

PinkyWay
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