A Very Powerful Manipulation Rule

There is one manipulation rule that I feel is not emphasized enough (for me at least) and that is:

P ⇒ Q       ⇔     ¬P ∨ Q

To me, this is one of most powerful manipulation rules that is extremely important to know so I didn't just want to memorize it. I wanted to understand it.

At first, It was kind of hard to wrap my head around the idea that "If P is true, then Q must be true" is the exact same as "P is not true or Q is true." When trying to connect these two statements using English only, it felt kind of hard to visualize, for me at least.

So I ended up working out several different ways to try to grasp this idea logically.

First, I did the truth tables:

P	Q	P ⇒ Q	¬P ∨ Q
T	T	T	T
T	F	F	F
F	T	T	T
F	F	T	T

After I did the truth tables, it was clearly evident that the implication and the logical disjunction statements were identical (according to the truth table). However, in terms of wording and a way of making English sense of it in my head, I was still not fully understanding this. So I continued:

Second, I did the Venn Diagram for both:

For P ⇒ Q : If P is true, then Q must be true, meaning there cannot be anything in the set of P where Q is not true, hence:

(Where P is the antecedent, Q is the consequent, and U is the universe. White is unoccupied)

For ¬P ∨ Q : Everything that is not P can be occupied and all of Q can be occupied, hence:

(Where P and Q are both predicates and U is the universe. White is unoccupied)

After creating the Venn Diagrams for both. It was starting to make a lot more sense. The Venn Diagrams are exactly the same. Since P implies Q means that "If P is true, then Q must be true" which means that there can't be any P's that aren't Q, then that would mean anything that's not P is fine. Also, if it's not P, then it doesn't matter what Q is for the implication as a whole to be true; vacuously. This is making a lot of sense. However, I still wanted to justify this some more!

Finally, I found negating the implication helped finalize my understanding of this manipulation rule:

First, "If P is true, then Q must be true" may be reworded with sets: "Every D that is also a P is also a Q" which would symbolically be:

∀x ∈ D, P(x) ⇒ Q(x)

So, if this was negated, the English would be "There exists a D that is also a P but is not Q." Symbolically:

∃x ∈ D, P(x) ∧ ¬Q(x)

So, I tried to negate the negation, to get back the original implication using De Morgan's Law:

¬(∃x ∈ D, P(x) ∧ ¬Q(x))

∀x ∈ D, ¬(P(x) ∧ ¬Q(x))

∀x ∈ D, (¬P ∨ Q)

Here I stopped and I didn't bother to change the disjunction back to an implication because I now saw that they are truly equivalent. This may have seemed like a roundabout way to get this the implication and disjunction stuck in your head, but this really helped me out! I thought to myself "I think I fully understand this concept now!"