Simplify the following Boolean function: ab\'(ab\'+b\'c) Here is my attempt, ple

ID: 1814349 • Letter: S

Question

Simplify the following Boolean function: ab'(ab'+b'c)

Here is my attempt, please explain where my logic is incorrect.

=ab'(ab' + b'c)

=(ab')(ab') + (ab')(b'c) ---> By distribution

=(ab') + (ab')(b'c) ---> (ab')(ab') = (ab') by absorption

=(ab') + ((ab'(c+c'))(b'c) ---> since (ab') = (ab'1), and since c+c' = 1

=(ab') + (ab'c+ab'c')(b'c) ---> By distribution

=(ab') + (ab'c)(b'c) + (ab'c')(b'c) ---> By distribution again

=(ab') + (ab'c) + (ab'c'c) ---> (ab'c)(b'c) = (ab'c) by absorption, and (ab'c')(b'c) = (ab'c'c) also by absorption

=(ab') + (ab'c) ---> since cc' = 0

=ab'(1+c) ---> factoring out (ab')

=ab' ---> since (1+c) = c, and since ab'1 = ab'

ab'(ab' + b'c)

=(ab')(ab') + (ab')(b'c) ---> By distribution

=(ab') + (ab' *b'c) ....> b' * b' =b'

= ab' + ab'c

= ab'( 1+c)

= ab'

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