Example $11\mathrm{.}57$ very quickly shows that the set of positive integer divisors of $30$ forms a Boolean

algebra. It actually takes quite some $($easy but tedious$)$ work to verify that it satisfies all of the

laws, and this is a subtle process.

Take the set S $=\{0,1,$a$,$b$,$c$,$d$\}$ with the complements defined as:

$,,($and for any x to get the other $3$ complements$)$

Define $+$ via: a$+$b $=$ b$,$ a$+$c$=$a$+$d$=$b$+$c$=$b$+$d$=1,$ and x$+$y$=$y$+$x$,0+$x $=$ x$,$ x$+$x $=$ x$,$ for any x$,$y to get the

others

Define $$ via: a$$b $=$ a$,$ a$$c$=$a$$d$=$b$$c$=$b$$d$=0,$ and x$$y$=$y$$x$,1$x $=$ x$,$ x$$x $=$ x$,$ for any x$,$y to get the others

Show that this is not a boolean algebra, by pointing out which part of Definition $11\mathrm{.}56$ fails