Logic – Relation between Implication and Disjunction

logic

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

Best Answer

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.

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