Prove De Morgan's second law, ~(AB) = ~A v ~B, by applying De Morgan's first law to ~A and ~B. I think I know how to do it but I just need some clarification to justify my answer... Thanks
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Okay, then post your solution to see if it's correct. P.S.: that should be $\displaystyle \sim(A\wedge B).$
I don't understand how this can be used to prove anything since $\displaystyle \sim(A\wedge B) \not{\equiv} \sim A ~\wedge \sim B$.
similar problem done here
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