Boolean algebra simplify the expression 3

Question

Simplify the expression using the rules in boolean algebra: ae + abc'd + bc'e'

Here's a list of the boolean rules

I got no solution to this problem. Anyone got any suggestions?

Answer 1

$$ae + abc'd + bc'e' \overset{Adjacency}= ae + abc'de + abc'de' + bc'e' \overset{Absorption \ x \ 2}=ae + bc'e'$$

In other words, the trick is to split the $abc'd$ term into two cases: one where $e$ is true, and one where $e$ is false ... and both of these terms are covered (i.e. get 'absorbed') by the existing terms respectively.

Absorption is B16 by the way.... and I don't see Adjacency on the list, but that's easily derived:

$$abc'd = abc'd1 = abc'd(e + e') = abc'de + abc'de'$$