Basic discrete math question regarding translation of logic ↔ English I just started Discrete Mathematics, and am having a little bit of trouble in understanding the conversions of English ↔ logic.

*

*$p$: "you get an A on the final exam."

*$q$: "you do every exercise in the book."

*$r$: "you get an A in the class."


Problem 1
Convert to logic using $p$, $q$, and $r$:

You get an A on the final, but you don't do every exercise in the book.  Nevertheless, you get an A in the class.


What's screwing me up is the word “nevertheless.”  At first I wrote $p \land \lnot q \land r$, then I thought about it and changed it to $(p \land \lnot q) \to r$, and somehow that seems like a better answer, but I have no idea why.
What I find hard to understand is when a sentence requires a conditional ($\to$).  Does the translation for the above sentences need a conditional?  There was a question earlier in the homework that said to translate $p \lor q \lor r$ to English, but how do you translate a bunch of variables with no conditional into an English sentence that makes sense?
 A: You can take the word "nevertheless" as a clue indicating that what follows isn't implied by and clearly doesn't depend on, the preceding knowledge. "Nevertheless" could easily be replaced by the word "but" and still convey the same meaning, and as you know, we translate "but" as meaning "and": 
"You get an A on the final, but you don't do every exercise in the book. But, you get an A in the class." This says pretty much what the original statement says.
If there were an implication (conditional) involved in the sequence of statements, you'd see something like "...Therefore, ..." or "As a consequence, ...".  
So the correct translation here would indeed be $$(p \land \lnot q) \land r,\quad\text{or simply}\quad p \land \lnot q \land r$$
A: You only use a conditional when you are given a conditional (causal statement), express or implied. A conditional such as $A\rightarrow B$ is very powerful because it asserts that there is a logical relationship between A and B, namely that if A is true, B must follow. 
You do not have a conditional relationship here; you don't know that r followed all the time from $p \wedge \neg q$. It is certainly possible that even if you get an A on the final but don't do all the exercises, you do NOT get an A in the class. As such I would lean towards your original thought, $p\wedge\neg q \wedge r$.
