Proof: We shall prove parts $($a$)$ and $($b$)$ and leave the remaining parts for the exercises.

a$)$ To prove that AsubeC, we need to verify that for all xinU, if xinA then xinC. We start

with an element $x$ from $A.$ Since AsubeB,xinA implies xinB. Then with BsubeC,xinB

implies xinC. So xinA implies xinC $($by the Law of the Syllogism $-$ Rule $2$ in Table

$2\mathrm{.}19-$since xinA,xinB, and xinC are statements$),$ and AsubeC.

b$)$ Since AsubB, if xinA then xinB. With BsubeC, it then follows that xinC, so AsubeC.

However, AsubB$=>$ there exists an element binB such that $b!$inA. Because BsubeC,

binB$=>$binC. Thus AsubeC and there exists an element binC with $b!$inA, so

AsubC.