Why is $P \to Q \equiv \neg P \vee Q$?

Question

By truth table, we know that $P \to Q$ is equivalent to $\neg P \vee Q$.

But I'm trying to understand why this work? How can connective "or" be implication. I tried some examples but I still can't find how they're the same.

Answer 1

The idea is this: The statement "$P$ implies $Q$" is true if $P$ is false, because the statement only tells us something if $P$ is true. So if $P$ is false, then the statement becomes vacuously true. On the other hand, if $P$ is true, then the statement "$P$ implies $Q$" can only be true if $Q$ is also true, because otherwise the statement says that something true implies something false. So "$P$ implies $Q$" is true if and only if $P$ is false or $Q$ is true, that is if "not $P$ or $Q$".

Answer 2

A slightly different point of view would consist in noting that saying $P\rightarrow Q$ is not true. Now an implication in not true if there exists one counter-example, i.e. if $ P \wedge \neg Q $ is true. Then use De Morgan's laws to obtain $\neg P \vee Q$.

Answer 3

If the implication, $P\to Q$, holds, then we know that $Q$ is true whenever $P$ is true. Thus either $P$ is false or $Q$ is true; that is $\neg P\vee Q$ .

If the disjunction, $\neg P\vee Q$, holds, then we know that if $P$ is true then $Q$ must be too; that is $P\to Q$ .

Therefore $\qquad P\to Q \iff \neg P \vee Q\qquad$ Q.E.D.

Answer 4

I'm a little late to the party, but I thought I would share two things for the OP that may be of use (surprised they have not been addressed already in the answers):

1. The logical reason for why one statement is equivalent to another may be rigorously examined by outlining a truth table and showing that the truth values for the two statements are the same regardless of the truth values of the entries.
2. A linguistic consideration is probably the most likely candidate to shed intuitive reason as to why two statements are equivalent.

For (1), consider the following truth table: $$ \boxed{ \begin{array}{c|c|c|c} P & Q & \neg P & \neg P\lor Q & P\to Q \\ \hline T & T & F & T & T \\ T & F & F & F & F \\ F & T & T & T & T \\ F & F & T & T & T \end{array}} $$ Of course, the above truth table is only likely to help if you agree on the truth values for $P\to Q$ (that is, it assumes you are comfortable with the fact that $P\to Q$ is always true except for when $P$ is true and $Q$ is false). Regardless, you know beyond the shadow of a doubt now that $P\to Q\equiv\neg P\lor Q$; however, the intuition behind this equivalence is still lacking.

For (2), consider a simple "if-then" expression in English: "If my girlfriend dumps me, then my heart will be broken." We can put this expression in the form $P\to Q$:

• $P:$ My girlfriend dumps me.
• $Q:$ My heart will be broken.

Now think about what $\neg P\lor Q$ actually means: "Either my girlfriend does not dump me or my heart will be broken." Kind of makes sense now how $\neg P\lor Q\equiv P\to Q$ doesn't it?

There are many other ways to try to explain this equivalence away, but I think a truth table argument is good for indisputable proof and then a linguistic example to provide that much sought after intuition.

Answer 5

It is easier to see why $P\implies Q\equiv \neg [P \land \neg Q]$.

Consider the statement: If it raining, then it is cloudy.

Symbolically:

$Raining \implies Cloudy$

Many beginners confuse this with cloudiness somehow causing rain, or rain casusing cloudiness. Neither is the case. There is no suggestion of causality in this statement. It just means that it is never both raining and not cloudy.

$\neg [Raining \land \neg Cloudy]$

Answer 6

An other approche (I'll use set theory). Let $A,B$ two set and suppose $A\subset B$ and let $x\in A$. Then, by hypothesis, $x\notin A\cap B^c=\emptyset$. Therefore, $x\in (A\cap B^c)^c=A^c\cup B$. Reciprocally, if $x\in A^c\cup B$ then $x\notin A\cap B^c$, then if $x\in A$, then $x\notin B^c$ and thus $x\in B$. That mean $A\subset B$. If $x\in B^c$ then $x\notin A$ and thus $B\supset A$.

I hope it will help :-)