# De Morgan's second law Proof

• May 10th 2008, 04:02 PM
jlopez
De 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 PM
Krizalid
Okay, then post your solution to see if it's correct.

P.S.: that should be $\sim(A\wedge B).$
• May 10th 2008, 05:06 PM
angel.white
I don't understand how this can be used to prove anything since $\sim(A\wedge B) \not{\equiv} \sim A ~\wedge \sim B$.
• January 23rd 2010, 08:07 PM
Jhevon
similar problem done here