Using the Boolean identities, simplify the following expressions:

Question

I am really not sure of my answer, could someone check it for me? Thank you

(rs'+ q't) (qr' + st')
((rs' + q')(rs' + t))((st' + q)(st' + r'))                      distributive law
(q'+ r)(q'+ s')(t + r)(t + s')(q + s)(q + t)(r' + s)(r' + t')   distributive law
(q' + s')(q + s)(t + r)(t' + r')(q' + r)(t + s')(q + t)(r'+ s)  commutative law
0 0 (q' + r)(t + s')(q + t)(r'+ s)                              inverse law
0                                                               null law

Answer 1

Small mistake on line 3:

$(q'+ r)(q'+ s')(t + r)(t + s')(q + s)(q + t)(r' + s)(r' + t')$

That should be:

$(q'+ r)(q'+ s')(t + r)(t + s')(q + s)(q + t\color{red}')(r' + s)(r' + t')$

Big mistake on line 5:

$(q' + s')(q + s) \not = 0$

The inverse of $q + s$ is $q's'$, not $q' + s'$

Also, I would start out quite differently:

$ (rs'+ q't) (qr' + st') \overset{Distributive \ Law}= $

$ (rs'+ q't) qr' + (rs'+ q't)st' \overset{Distributive \ Law}= $

$rs'qr'+rs'st'+q'tqr'+q'tst' = ...$