Just checking if I'm right with these? Given F=A'B+(C'+E)(D+F'), use de Morgan's theorem to find F' I got F=A+B'(CE')+(D'F)???? Simplify the following: {[(AB)'C]'D}' I got A'+(B'+C')D Pretty tough for me
ckaiser813, You need to read this thread and download this program. http://forum.allaboutcircuits.com/showthread.php?t=22347&highlight=BOOLEAN When you do plug in the Boolean expression, A'B+(C'+E)(D+F'), into the program, you will obtain the following minterms: 0 2 4-7 10 14-32 34 36-39 42 46-48 50 52-55 58 62-63 The negative of that expression should have the missing minterms of the above expression, which are: 1 3 8-9 11-13 33 35 40-41 43-45 49 51 56-57 59-61 Since your answer, A+B'(CE')+(D'F) does not these minterms, that is a sure indication that your result is wrong. Try again and use the program to check your new result. The program gives the minterms for ((AB)'C)'D as 1 5 9 13 15 Your answer, A'+(B'+C')D, has minterms 0-7 9 11 13 , so again you have the wrong answer. Get the idea? Ratch
Thanks for your help, I'd never used that type of program but it help me understand what to look for. thanks again!