Propositonal equivalence and compound proposition 
Without using truth tables, show that the statements 
  
  
*
  
*‘If you did all problems in the book, attended all lectures and completed all assignments, then you will get an A in Discrete Math’ and 
  
*‘If you did all assignments but did not get an A in Discrete Math, then you either did not do all problems in the book or did not attend all lectures’ 
  
  
  are logically equivalent.  
Hint: first translate the given sentences to compound propositions.

My solution
Let A be the proposition ‘you did all problems in the book,’
    B be the proposition ‘you attended all lectures,’
    C be the proposition ‘you completed all assignments’ and
    D be the proposition ‘you got an A in Discrete Maths.’
Therefore, ‘If you did all problems in the book, attended all lectures and completed all assignments, then you will get an A in Discrete Math’ translates to (A AND B AND C) IMPLIES D and ‘If you did all assignments but did not get an A in Discrete Math, then you either did not do all problems in the book or did not attend all lectures’ translates to [A AND (NOT D)] IMPLIES [(NOT)B OR (NOT)C].
(A AND B AND C) IMPLIES D
= [NOT (A AND B AND C)] OR D
= [ (NOT A) OR (NOT B) OR (NOT C)] OR D
= [(NOT A) OR D] OR [(NOT B) OR (NOT C)]
= [NOT (A AND (NOT D))] OR [(NOT B) OR (NOT C)]
= [A AND (NOT D)] IMPLIES [(NOT)B OR (NOT)C]
Do you think my answer is correct?
 A: There's a small mistake there,  “If you did all assignments but did not get an A in Discrete Math, then you either did not do all problems in the book or did not attend all lectures” translates to “$\small\rm (C\land \lnot\,  D)\Rightarrow(\lnot\,  A\lor\lnot\,  B)$” rather than “$\small\rm (A\land \lnot\,  D)\Rightarrow(\lnot\,  C\lor\lnot\,  B)$”, but this wont have any consequence on the result since interchanging $\small\rm A,B,$ and $\small\rm C$ will yield basically the same thing. In any case, I changed your proof so that it yields the requested result:
$$\small\begin{align}
\mathrm{(A \land  B \land  C) \Rightarrow  D}
 & \equiv \mathrm{ \big(\lnot\,   (A \land  B \land  C)\big) \lor  D}
\\ 
& 
\equiv \mathrm{ \big( (\lnot\,   A) \lor  (\lnot\,   B) \lor  (\lnot\,   C)\big) \lor  D}
\\
& \rm
\equiv  \big((\lnot\,   C) \lor  D\big) \lor  \big((\lnot\,   A) \lor  (\lnot\,   B)\big) 
\\
& 
\equiv  \mathrm{\big(\lnot\,   (C \land  (\lnot\,   D))\big) \lor  \big((\lnot\,   A) \lor  (\lnot\,   B)\big)} 
\\
&
\equiv\mathrm{  \big(C \land  \lnot\,   D\big)\Rightarrow  \big(\lnot\,  A \lor  \lnot\,  B\big)}.\tag*{$\square$}
\end{align}$$
