Interpreting resolution rule of inference I know that the the resolution rule of inference states that 

$(p \lor r)\wedge (q \lor \lnot r) \to (p \lor q)$

Based on this, my textbook says that below statement is true:

$(p \vee q)$ is satisfiable if and only if $(p \lor r)\wedge (q \lor \lnot r)$ is satisfiable.

However if we look at the resolution rule, it says LHS implies RHS. Thus even if LHS is false, RHS can be true. Doesn't that mean, RHS can be satisfied even if LHS is unsatisfiable and thus not only when LHS is satisfiable?
 A: Note that the claim is just that they're both satisfiable, not necessarily in the very same models (assignments of truth values ot variables). In particular, there's no claim that the converse is a valid inference rule -- it isn't, as the converse of the implication is not valid.
Because the given formula
$$
\vdash (p \lor r)\wedge (q \lor \lnot r) \to (p \lor q)\tag{1}
$$
is valid (a tautology, provable), certainly it's clear that if $(p \lor r)\wedge (q \lor \lnot r)$ is satisfiable then so is $(p \lor q)$. In fact, any assignment of truth values to propositional variables that satisfies the former, also satisfies the latter, because it has to assign true to at least one of $p, q$.
Now suppose $(p \lor q)$ is satisfiable. Then under some assignment $v$ of truth values to variables, $v(p) = 1$ or $v(q) = 1 = true$ — without loss of generality, say $v(p) = 1$. Let $v'$ agree with $v$ on all variables except possibly $r$, where $v'(r) = 0 = false$. Then 
$$
v'((p \lor r)\wedge (q \lor \lnot r)) = 1,
$$
so $(p \lor r)\wedge (q \lor \lnot r)$ is satisfiable too.
A: The second statement (with the "if and only if") is false (consider what would happen if we had $p$ and $r$ true and $q$ false). The two formulae are not equivalent.
