$[12$ marks$]$ This question involves Boolean aigebra. Solutions must show sequential equivalent Boolean expressions. Moving from one expression to the next in the proof must include a justification requiring no more than two rules.$|$You may only use the the Boolean laws covered in the slides, and the distributive law for OR that was part of Tutorial $1,$ as justification.

$($a$)$ Prove:

$(not{A}^{{?}^{?}}not{B}^{{?}^{?}}$notC$)vv(A^{{?}^{?}}not{B}^{{?}^{?}}$notC$)vv(not{A}^{{?}^{?}}not{B}^{{?}^{?}}C)=not{B}^{{?}^{?}}$not$(A^{{?}^{?}}C)$

$($b$)$ Simplify the following Boolean expression as much as possible. An expression that is simpler than another will have fewer Boolean operators. Parenthesis are not Boolean operators.

$A^{{?}^{?}}Bvv((AvvC{)}^{{?}^{?}}{B}^{{?}^{?}}$not$(AvvC))vvnot(notAvvB)$

$($c$)$ Simplify the following Boolean expression as much as possible. An expression that is simpler than another will have fewer Boolean operators. Parenthesis are not Boolean operators.

$(A^{{?}^{?}}not{B}^{{?}^{?}}$notC$)vv(A^{{?}^{?}}not{B}^{{?}^{?}}C)vv(A^{{?}^{?}}$notD$)vv(not{A}^{{?}^{?}}$notB$)vvnot(AvvD)$