Relation between Implication and Disjunction I hope that someone help me with an intuition or a good explanation for why does Implication relate to Disjunction in Mathematical Logic.
The specific property I'm asking about is :  
                                (P → Q) ↔ (¬P ∨ Q)

Also, while I was searching on Wikipedia I found some Venn Diagrams explaining with visuals the Truth Table which I used to Proof the mentioned property, but it didn't make sense for me, eventually the symbolic proof doesn't build a good intuition in our minds.
Thank you for your time, and I apologize if this question was previously asked in way or another, I couldn't find any, this is my first post here :).
The link of the Wikipedia Article : https://en.wikipedia.org/wiki/Logical_connective
 A: In classical logic, the logical connectives are truth-functional, i.e. defined by their truth-table.
With them, we may easily verify that the two formulae :

$P \to Q$ and $\lnot P \lor Q$

are equivalent.

If you are not "at ease" with this result, you are not alone: in Intuitionsitic Logic this result is not provable.
In effect, this logic does not "agree" on the truth-functional definition of the connectives.

In the context of "classical logic" the best "intuition I can suggest you is to reflect how the $\to$ connective is used in mathematical reasoning by way of modus ponens rule of inference.


This rule licenses us to infer $B$ from $A$ and $A \to B$.


We "apply" it asserting the premises :$A \to B$ and $A$.
The first assertion excludes the case when $A$ is true and $B$ false, while the second assertion excludes the two cases where $A$ is false.
Thus, the truth-table fo $\to$ leaves us only one possibility : $B$ true, and this is what we need in order to conclude the proof.
The same happens with the "disjunction version" of the same rule :


that licenses us to infer $B$ from $A$ and $\lnot A \lor B$.


In words : if we know that at least one between $\lnot A$ and $B$ holds and if we know that $A$ holds (i.e. $\lnot A$ does not), we have to conclude that $B$ must hold.
A: Note: $\lnot p\lor p\equiv T$. So  $p\rightarrow q$ necessitates $\lnot p\lor q$.


*

*Edit-The statement $\lnot p\lor p$ is a tautology (i.e. always true).Take few examples,"Either it is raining or not raining","Either I wear white shirt or I do not", "Either I am on MSE or I am not", etc. So, plugging in your conditional $p\rightarrow q$ (which means $q$ happens whenever $p$ happens) in $\lnot p\lor p$, one gets $\lnot p\lor q$.

A: if P, then Q (that is, P implies Q).
You want to know if the implication is true. If P is false, then it does not matter, if Q true or false. Overall, the statement is true.
For P to imply Q, when P is true, Q must be true. Therefore, it is necessary for Q to be true, when P is true, for P to imply Q.
Thus for the implication to be true, either P is false or Q has to be true. 
Example: Let X be (0<1), a true statement. Let Y be (0<2), a true statement. 
Then "not X" is (0>=1) and "not Y" is (0>=2).
X implies Y versus (not X or Y)
(0 >= 1) implies (0 < 2): (strange, but) true.
We have not X, so (not X or Y) is true
(0 >= 1) implies (0 >= 2): (strange, but) true.
We have not X, so (not X or Y) is true
(0 < 1) implies (0 >= 2). The implication is false.
Here we have X and not Y, so (not X or Y) is false
(0 < 1) implies (0 < 2). The implication is true.
We have X and Y, so (not X or Y) is true. 
