Equivalence of $P\rightarrow Q$ and $\lnot P\lor Q$

Question

How do we explain the logical equivalence $$(P\rightarrow Q ) \equiv [(\neg P)\; \vee \; Q]$$ and if possible could you please give an example illustrating this equivalence. Thanks alot !!

Answer 1

By definition, the implication $P\rightarrow Q$ is true whenever $P$ is false, or whenever $Q$ is true: $$\lnot P\lor Q$$

Alternatively, you can compare the truth-table for $P\rightarrow Q$ side-by-side with the truth-table for $\lnot P \lor Q$ and see that under the value in their rightmost columns, we have agreement for each and every truth-value assignment. Hence, they are logically equivalent.

Answer 2

Here's an illustration:

Let $P$ be "You get a good grade on exam" and $Q$ be "I'll buy you an ice-cream". So $P $$\to$$ Q$ reads "If you get a good grade on exam, then I'll buy you an ice-cream". But that is the same as to say "Either you don't get a good grade on exam, or I'll buy you an ice-cream", which is exactly $\neg P \vee Q$. I think you can make your own illustrations similar to this argument, they all make sense in real-life language.

Answer 3

By definition $(P\rightarrow Q)$ is true whenever $P$ is false, or whenever $Q$ is true. Therefore : $$(P\rightarrow Q) \equiv(( \lnot P)\lor Q)$$

Answer 4

You can always use truth tables of both logical statements because it is only four combinations of truth - > (T,T) (T,F) (F,T) and (F F).

Answer 5

Instead of using a truth table or a series of logical equivalences, I'd like to try a more intuitive method for you.

Try drawing a Euler diagram of $P\to Q$, as in, a small circle $P$ encompassed by a lager circle $Q$.

Now you can see, no matter where you point on this diagram, it is true you will point at $\lnot P$ or you will point within $Q$, or both.