"If p, then q" = "q unless ~p"? Can anybody please explain to me how "if $p$, then $q$" is logically equivalent to "$q$ unless ~$p$" ?
My problem is with all truth possibilities of $p$ and $q$ except for when both are true.
It seems in those situations they would not have the same logical meaning.
For e.g: "if $p$, then $q$" is true if $p$ is false and $q$ is true. However, shouldn't this possible combination render "$q$ unless ~$p$" false?
Feel free to use truth tables or whatever method that would help in explaining.
 A: Logicians tend to translate "$p$ unless $q$" as 
$$
p \vee q
$$
which is equivalent to both
$$
\sim p \to q
$$
and
$$
\sim q \to p
$$
You can get a reading of the English sentence "$p$, unless $q$" on which it's making a kind of conditional assertion.  It's asserting that $p$, but only so long as $q$ isn't true.  If $q$ turns out to be true, then an assertion of "$p$, unless $q$" simply falls silent on the question of whether or not $p$.  If you're hearing this reading, then the English sentence "$p$ unless $q$" would just be ruling out the possibility that both $p$ and $q$ are false---and that's just the possibility that "$p \vee q$" rules out.
$$
\begin{array}{c  c | c}
p & q & p \vee q \\\hline
T & T & T\\
T & F & T\\
F & T & T\\
F & F & F
\end{array}
$$
But, you can get another reading of the sentence "$p$, unless $q$", on which it's not merely saying that $p$ is true, given that $q$ is false.  You can get a reading on which it's additionally saying that $p$ will be false if $q$ is true.
You're probably hearing that second reading of "$p$, unless $q$".  In that case, you'd want to translate that into propositional logic with "$\sim p \equiv q$"; or, equivalently, "$p \equiv \sim q$"; or, equivalently, "$\sim(p \equiv q)$".
$$
\begin{array}{c  c | c | c | c}
p & q & \sim p \equiv q & p \equiv \sim q & \sim(p \equiv q) \\\hline
T & T & F & F & F\\
T & F & T & T & T\\
F & T & T & T & T\\
F & F & F & F & F
\end{array}
$$
Which of these translations is correct isn't properly a question for logic to settle.  It's a question for linguistics to settle.  I generally explain this to my students and tell them that they should just get used to the fact that logicians tend to use "$p$, unless $q$" in the first way.
A: Natural languages are ambiguous, so if you want to check a logical equivalence you should "translate" your propositions to the language of some form of logic (propositional logic, predicate logic, ...).  The title of your question is mistaken but then in the description you write it right. "If $P$ then $Q$" is equivalent to "$Q$ unless $\lnot P$" because to make the implication $P \implies Q$ be true when $Q$ is false it must be the case that $P$ is also false (which meand $\lnot P$ is true), otherwise the implication doesn't hold. $Q$ unless $\lnot P$ should be "translated" as $Q \lor \lnot P$ 
