Using the resolution proof system so I can't figure out something when using the resolution proof system. 
We are given: {¬a → ¬b, b} on a
what I can't figure out is why: ¬a → ¬b becomes a ∨ ¬b
I understand it becomes an ∨, but why does ¬a get changed it to a?
Thanks.
 A: 
You have to check that $\lnot a → \lnot b$ is equivalent to $a \lor \lnot b$ via truth table.

I'll details the four cases entering the truth-value for $a$ and $b$.
(i) : $T-T$
In this case $\lnot a → \lnot b$ is $F \rightarrow F$ which, by truth-functional definition of $\rightarrow$ is $T$.
While $a \lor \lnot b$ is $T \lor F$, which also is $T$.
(ii) : $T-F$
In this case $\lnot a → \lnot b$ is $F \rightarrow T$, which is $T$.
While $a \lor \lnot b$ is $T \lor T$, which again is $T$.
(iii) : $F-T$
In this case $\lnot a → \lnot b$ is $T \rightarrow F$, which is $F$.
But $a \lor \lnot b$ is $F \lor F$, which is $F$ also.
Finally :
(iv) : $F-F$
In this case $\lnot a → \lnot b$ is $T \rightarrow T$, which is $T$.
While $a \lor \lnot b$ is $F \lor T$, which is $T$.
Having covered all possible cases, we can see that in each case the two formulae has the same truth value; thus, they are equivalent.
Comment
The "brief" explanation is that 

$a → b$ is equivalent to $\lnot a \lor b$. 

Thus, starting from $\lnot a → \lnot b$ we get : $\lnot \lnot a \lor \lnot b$.
Then we apply double negation to rewrite it as : 


$a \lor \lnot b$.


See Conjunctive normal form for the explanation of the method, which amounts to :

Every propositional formula can be converted into an equivalent formula that is in CNF. This transformation is based on rules about logical equivalences: the double negative law, De Morgan's laws, and the distributive law.

See example regarding : Convert to negation normal form. 
A: Given a formal language and being $\alpha$ and $\beta$ two formulas, the notation $\alpha \vee \beta$ is normally presented as a convention for writting $¬\alpha \rightarrow \beta $. If you define $\alpha \equiv a$ and $\beta \equiv ¬ b$, then $\ ¬a \rightarrow ¬b$ is equivalent to $\ ¬\alpha \rightarrow \beta$ is equivalent to $\alpha \vee \beta$ is equivalent to $a \vee ¬b$.
Also, for convince yourself that $\alpha \vee \beta$  and $¬\alpha \rightarrow \beta$ are equivalent formulas and that I'm not cheating you, you can compare the truth table of the first one with the truth table of the second one and verify that they are the same one.
