Yes. Double NOT the function (this doesn't change the function), so: f=//(/cd+/ab+/a/c+a/bcd). Now think about DeMorgan's theorems. Have a go and post up your efforts, I'll give you some feedback. Dave
I did the boolean algebra again this time using k maps and realized I had made a mistake in reducing my equation. The reduced one now is /CD+/AB+/A/C+A/BD and I don't think this can be reduced further. Can it?
Sorry I missed this one. Feel free to PM to remind me in future. Without your working out I cannot give you any feedback on your answer, but I simplified it to: f = (ABCD)' (Note I use ' to denote NOT/INVERTED) I used the two techniques I told you about above. How? f = (CD)' + (AB)' + (A'B') + (AB'CD) Double NOT (this doesn't change the function): f = ((CD)' + (AB)' + A'B' + AB'CD)'' Take out a common factor in the last two terms: f = ((CD)' + (AB)' + B'( A' + ACD))'' Apply DeMorgan's theorem: f = ((CD)''(AB)''(B'(A' + ACD))')' (CD)'' = CD; (AB)'' = AB; so: f = (CDAB(B'(A' + ACD))')' Apply DeMorgan's theorem on (B'(A' + ACD))': f = (CDAB(B'' + (A' + ACD)'))' B'' = B Apply DeMorgan's theorem on (A' + ACD)': f = (CDAB(B + (A''(ACD)')))' A'' = A Apply DeMorgan's theorem on (ACD)': f = (CDAB(B + (A(A' + C' + D'))))' Multiply out (A(A' + C' + D'): f = (CDAB(B + (AA' + AC' + AD')))' AA' = 0 f = (CDAB(B + AC' + AD'))' Multiply everything out: f = (CDABB + CDABAC' + CDABAD')' BB = B; CC' = 0; DD' = 0; therefore f = (CDAB)' = (ABCD)' Not easy eh?! Dave