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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- May 10th 2008, 04:02 PMjlopezDe Morgan's second law Proof
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 - May 10th 2008, 05:01 PMKrizalid
Okay, then post your solution to see if it's correct.

P.S.: that should be $\displaystyle \sim(A\wedge B).$ - May 10th 2008, 05:06 PMangel.white
I don't understand how this can be used to prove anything since $\displaystyle \sim(A\wedge B) \not{\equiv} \sim A ~\wedge \sim B$.

- Jan 23rd 2010, 08:07 PMJhevon
similar problem done here