# Question about logical implication $P\to Q$ [duplicate]

Having come across mathematical logic, a question suddenly came into my mind. We commonly know that the truth value of $P\to Q$ given as:

$\begin{matrix} P&Q&P \Rightarrow Q \\ T&T&T\\ T&F&F\\ F&T&T\\ F&F&T \end{matrix}$

I do not understand how $P\to Q$ holds when P is false. For example, let me propose a statement:

Let $n$ be a nonzero real number.

P: $n$ is a rational number

Q: $n\cdot0=k$, where $k$ is a nonzero real number.

We obviously know that Q is a false statement. Hence, I omit the case when Q is true. $\begin{matrix} P&Q&P\Rightarrow Q \\ T&F&F\\ F&F&T \end{matrix}$

P True: $n$ is a rational number;

P False: $n$ is not a rational number; $n$ is an irrational number.

How is it that $P\to Q$ holds true when P is false?

## marked as duplicate by Gyu Eun Lee, MJD, ronno, Claude Leibovici, JohnDNov 23 '14 at 5:14

• What does "P Q P->Q T T T T F F F T T F F F" mean? – GFauxPas Nov 23 '14 at 1:13
• @GFauxPas - It's a bad formatted truth table. – Nagase Nov 23 '14 at 1:15
• Also math.stackexchange.com/questions/100286/… and probably a dozen others. – MJD Nov 23 '14 at 4:25
• What if answers to the previous question are felt to be entirely unsatisfactory? What is a struggling student to do other than repost a similar question? – Dan Christensen Nov 23 '14 at 6:40
• In that case the struggling student should explain why the eight previous answers are unsatisfactory. – MJD Nov 24 '14 at 6:11

When we say, if it is raining, then it is cloudy, there is often the erroneous suggestion that either cloudiness causes rain, or rain causes cloudiness. Neither is the case.

EDIT: It means only that, at the moment, it is not both raining and not cloudy. There is no suggestion of a causal relationship.

To answer your question, suppose it is not raining. Then it doesn't matter if it is cloudy or not, it cannot be both raining and not cloudy, i.e. it must be the case that, if it is raining, then it is cloudy.

Symbolically, we write:

Raining $$\implies$$ Cloudy

or equivalently

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

We define '$$\implies$$' as follows: $$[P \implies Q] \equiv \neg [P \land \neg Q]$$

With this in mind, the truth table for '$$\implies$$' makes perfect sense.

$$\begin{matrix} P&Q&P\implies Q&\neg [P \land \neg Q]\\ T&T&T&T\\ T&F&F&F\\ F&T&T&T\\ F&F&T&T \end{matrix}$$

Let's say there's a sign outside a basketball court that says: "If you're not wearing shoes, you cannot play basketball."

The negative of the antecedent is if I were wearing shoes. And if I were wearing shoes, I'm not violating what the sign says no matter whether or not I play basketball.

I only violate the sign if I'm not wearing shoes AND playing basketball (aka, it's false if T -> F).

Does that make sense?

• I have slightly edited the question with latex format. I hope it clarifies what I really wanted to ask. – James Chung Nov 23 '14 at 1:26
• My answer still stands. It may help to think about $P\Rightarrow Q$ as $\neg P \vee Q$. The implication statement doesn't apply if the antecedent is false, hence the implication still stands. – Harry Lime Nov 23 '14 at 1:30
• "If you have a cell phone, turn it off." You're in compliance if you have no cell phone. – Michael Hardy Nov 23 '14 at 1:38

Remember that $P \rightarrow Q$ has nothing to do with whether $Q$ is true. It only means:

If $P$ is true then $Q$ is true.

This always holds if P is false becasue if we assume that $P$ is true, while knowing that $P$ is false, we get:

$\neg P \wedge P$

And we can prove anything starting from this. So $Q$ is true.

In plain English, "$P\rightarrow Q$" is the same as "If P is true, then Q must be true". So, if you know that this kind of relation holds for two propositions, P and Q and someday i tell you that i observed P to be true and Q to be false in some situation, you will greet me in a second by "liar". But if i say, that i observed P is false and Q is true, then you won't call me a liar. Since you know that there can be another proposition which is making Q true or Q can be a universal truth. So, even if P is false and Q is true, it does not invalidate the relation "$P\rightarrow Q$". It violates the relation only when P is true and Q is false. That's why $P\rightarrow Q$ is also defined as, "it is not that P is true and Q is false". or logically "~P and Q".