I have been working on the following Boolean simplifications but I seem to be missing a theorem in my approach
1) AB + A'C + B'D + D'E + C'E Solution: AB + A'C + B'D + E
and
2) AD + BD + CD + A'B'C' Solution: D + A' B' C'
I notice that it seems that the solution seems to include something to the effect that if you have one equation that contains the notted version of existing variable that you can get ride of the existing, keeping the common. I know that may not make sense, but im a little lost on how to do these 2.
Thanks
1) AB + A'C + B'D + D'E + C'E Solution: AB + A'C + B'D + E
and
2) AD + BD + CD + A'B'C' Solution: D + A' B' C'
I notice that it seems that the solution seems to include something to the effect that if you have one equation that contains the notted version of existing variable that you can get ride of the existing, keeping the common. I know that may not make sense, but im a little lost on how to do these 2.
Thanks