Consider the following functions d $=$ c $($b $+$ a$)+$ a b e $=$ a b c $+$ b $($c $+$ a$)$ f $=$ b $($c $+$ a$)+$ a b $($a$)$ Express d in terms of e$,$ as well as the input variables. $[3$ points$]($b$)$ Express d in terms of f$,$ as well as the input variables. $[3$ points$]($c$)$ Express e in terms of d$,$ as well as the input variables. $[3$ points$]($d$)$ Express e in terms of f$,$ as well as the input variables. $[3$ points$]($e$)$ Express f in terms of d$,$ as well as the input variables. $[3$ points$]($f$)$ Express f in terms of e$,$ as well as the input variables. $[3$ points$]($g$)$ Define cost as the number of literals in the expression. $($A literal is an instance of a variable, whether negated or not.$)$ Assemble the lowest cost circuit, ensuring that it is combinational $($so all cycles are broken for each input combination$).$