if p then q unless r -- do i understand this? the statement is: if p, then q, unless r

I'm to convert it into a compound proposition.

I've reasoned that:

when p is true AND r is true, q is false
when p is true AND r is false, q is true
when p is false, we can't know if q is true

But, how to convert this into compound propositions?

$ (p \land \lnot r) \to q $ 
$ (p \land r) \to \lnot q $

How can we express both of these at the same time? or does each of them already contain the other?
Start with a truth table

 p | q | r || (p &amp; ~r) -&gt; q |(p &amp; r) -&gt; ~q  ||  p, then q, unless r
---|---|---||----------m----|--------m------||-----------------------
 T | T | T ||    F F   T    |   T    F  F   ||           ?
 T | T | F ||    T T   F    |   F    T  F   ||           ?
 T | F | T ||    F F   T    |   T    T  T   ||           ?
 T | F | F ||    T T   F    |   F    T  T   ||           ?
 F | T | T ||    F F   T    |   F    T  F   ||           ?
 F | T | F ||    F T   T    |   F    T  F   ||           ?
 F | F | T ||    F F   T    |   F    T  T   ||           ?
 F | F | F ||    F T   T    |   F    T  T   ||           ?


First conclusion $ (p \land \lnot r) \to q $ and $ (p \land r) \to \lnot q $ are not identical.

Then what is the truth table of "p, then q, unless r" ? 

That question you have to do yourself, and you will need to argue for each line without reference to the formula's

GOOD LUCK
It's $\lnot r \rightarrow (p \rightarrow q) $ I think