Equivalence of $a \rightarrow b$ and $\lnot a \vee b$ Is there a proof for the logical equivalence of $a \rightarrow b$ and $\lnot a \vee b$?
Try using a truth table comparing $p\rightarrow q$ with $\lnot p \lor q$.
That's as good a proof as you need.

$p$ | $q\,$ |$\;p\rightarrow q$ | $\lnot p \lor q$
T | T |$\quad$ T $\quad$|$\quad$ T 
T | F |$\quad$ F $\quad$|$\quad$ F
F | T |$\quad$ T $\quad$|$\quad$ T
F | F |$\quad$ T $\quad$|$\quad$ T

Both expressions evaluate false if and only if $\,p\,$ is true and $\,q\,$ is false.



For developing a bit of intuition regarding the conditional (logical implication): see, e.g.,
$(1)$ this post, and
$(2)$ this post.
Source 1 : http://plus.maths.org/content/os/issue36/features/nishiyama/index.
I understand this argument, but it does not feel intuitive. 


  We are given that $\color{green}{P \Longrightarrow Q}$.
  $\boxed{\text{Case 1 of 2 : } \color{green}{P \Longrightarrow Q} \text{ true}}$ Substitute the green into the tautology  $\color{green}{P} \vee \lnot P$ to get: $\color{green}{Q} \vee \lnot P$.
  $\boxed{\text{Case 2 of 2 : } \color{green}{P \Longrightarrow Q} \text{ false}}$ Since an argument is valid if the premises cannot all be true without the conclusion being true as well, thus this case means $[\text{$P$ is true $\&amp; \, Q$ is false]} \iff [\text{$\lnot P$ is false $\&amp; \, Q$ is false]} \iff [\text{$\lnot P \vee Q$ is false]}$. 
  
  Altogether, the 2 cases result in: 
  $\left\{ \begin{array}{rcl}
\color{green}{P \Longrightarrow Q} \text{ true $\equiv [\lnot$ P $\vee$ Q] is true} \\
\color{green}{P \Longrightarrow Q} \text{ false $\equiv [\lnot$ P $\vee$ Q] is false}
\end{array}\right.$.
  Thus, the Conditional Law has been proven.