Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicates:
In classical logic, why is (p -> q) True if both p and q are False?
Why an inconsistent formal system can prove everything?

I heard a professor of mathematics explaining that the following conslusions are true.

  1. $\textrm{False}\Rightarrow \textrm{True}$
  2. $\textrm{False}\Rightarrow \textrm{False}$

Could someone please explain to me why both conclusions are true? I am especially interested in the first case. I'd like to hear two explanations if possible: One for mathematicians and one for non-mathematicians. :-)

share|improve this question

marked as duplicate by amWhy, Asaf Karagila, mixedmath, J. M., t.b. Jul 27 '11 at 17:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3  
This question is relevant, in particular to your question $2$, but also generally (the reason $\text{False}\Rightarrow\text{True}$ and $\text{False}\Rightarrow\text{False}$ is by the very definition of $P\Rightarrow Q$ via a truth table). –  Zev Chonoles Jul 26 '11 at 2:08
1  
I think there are (at least) two related phenomena here: first that it is standard to define $\Rightarrow$ so that $\text{False} \Rightarrow \text{True}$, and second that if you have a contradiction in a logical system you can derive any statement from it. I am not sure these two are trivially equivalent. –  Qiaochu Yuan Jul 26 '11 at 2:14
2  
3  
Note that technically speaking, the conclusions are not both true. But the implications represented by 1. and 2. are both true. That is, if $A \implies B$ is true, that is not to say that B is therefore true (clearly, the conclusion of 2., "false", is not true; but $\text{False} \implies \text{False}$ is true –  amWhy Jul 26 '11 at 2:41
1  
Oh, oops. I only saw Ben's link. @Zev's link however would be an appropriate "close as duplicate" target. –  Willie Wong Jul 26 '11 at 9:57

5 Answers 5

up vote 13 down vote accepted

Here's one for the non-mathematician:

Let's say that I promise you $1,000,000 on the condition that pigs fly. We could say something like:

If pigs fly, then I will give you $1,000,000

Now, if pigs don't fly and I don't give you $1,000,000, you have no reason to complain. I haven't lied. This is your case #2.

If pigs don't fly yet I still give you $1,000,000 , (case 1) then I am just very generous, but I haven't contradicted myself or broken my word.

share|improve this answer
1  
As Zev and Q have mentioned, the mathematician's reason for this is either by (definition of implication) or (sufficiency of "or") –  The Chaz 2.0 Jul 26 '11 at 2:33

See http://www.nku.edu/~longa/classes/mat385_resources/docs/russellpope.html for Bertrand Russell's proof that 1=0 implies that he is the Pope.

share|improve this answer
1  
Delightful! (+1) –  The Chaz 2.0 Jul 26 '11 at 3:07

While the other answers have given mathematically adequate, I feel that perhaps I should explain why we define material implication this way.

First, let us consider the principle of modus ponens. If we know that $P$ is true, and we also know $P \to Q$, then it is reasonable that we may deduce $Q$ is also true. It is also reasonable to say that if it is possible to have $P$ true and $Q$ false at the same time, then we should have $P \to Q$ false. The question is how to define the truth value of $P \to Q$ so that the principle of modus ponens gives us sound deductions. It turns out there isn't a unique answer: the two ‘obvious’ properties of implication I just listed only force two of the entries in the truth table: $$\begin{array}{c|cc} & P \text{ false} & P \text{ true} \\ \hline \\ Q \text{ false} & ? & P \to Q \text{ false} \\ Q \text{ true} & ? & P \to Q \text{ true} \end{array}$$ It is possible to fill in the blanks in any way and still get an interpretation of $P \to Q$ which makes modus ponens valid. At least amongst mathematicians, it is conventional to choose the ‘weakest’ possible way to fill in the table, which is to say that $P \to Q$ is true whenever $P$ is false. This produces no logical inconsistencies and modus ponens is valid under this interpretation, so why not?

But actually, once we start adding other rules of inference, we rapidly find ourselves forced to conclude that $\to$ must be interpreted this way. Indeed, consider the following rules of inference:

  1. Monotonicity: From $P$, we may deduce $P$.
  2. Transitivity: If we may deduce $R$ from $Q$, and if we may deduce $Q$ from $P$, then from $P$ we may deduce $R$.
  3. Conditional proof: If we may deduce $R$ from $P$ and $Q$, then from $P$ we may deduce $Q \to R$.
  4. Negation introduction: If we may deduce $Q \to \bot$ from $P$, then from $P$ we may deduce $\lnot Q$. ($\bot$ is a proposition which is interpreted as ‘contradiction’. $\lnot$ is a logical operator which is interpreted as ‘not’.)
  5. Disjunction introduction: From $P$, we may deduce $P \lor Q$. ($\lor$ is a logical connective which is interpreted as ‘or’.)
  6. Modus tollendo ponens: From $P \lor Q$ and $\lnot P$, we may deduce $Q$.

I'm certain that everyone agrees that these rules of inference are entirely reasonable. But these already suffice to prove that ex falso quodlibet, that is, from $\bot$ we may deduce any proposition $Q$. Indeed:

  1. By monotonicity, we may deduce $\bot$ from $\bot$.
  2. By conditional proof, we may deduce $\bot \to \bot$ from $\bot$.
  3. By negation introduction, we may deduce $\lnot \bot$ from $\bot \to \bot$.
  4. By disjunction introduction, we may deduce $\bot \lor Q$ from $\bot$.
  5. By transitivity, we may deduce $\lnot \bot$ from $\bot$.
  6. By modus tollendo ponens and transitivity (a few times), we may deduce $Q$ from $\bot$.

We then may apply the principle of conditional proof again to deduce $\bot \to Q$ from any hypotheses whatsoever. So if we believe these rules of inference are valid, then we are forced to conclude that $P \to Q$ must be true whenever $P$ is false, no matter what $Q$ is.

share|improve this answer
    
I appreciate the formal approach! –  The Chaz 2.0 Jul 27 '11 at 19:54

The Chaz alluded, in a comment, that by sufficiency of "or" one can reason that both implications are true.

Note that $A \rightarrow B$ is equivalent to $\lnot A \lor B$ (you can confirm this using a truth table for each and convince yourself that they yield the same "output" - truth value - for any given combination of truth values of A and B.

Now, this is what always helped me when first learning logic:

Let the truth value of $A$ be false, of $B$, true. Then $A \implies B$ is an example of case 1 in which we have a false statement implying a true statement. Now, if we take the equivalence of $A \implies B$ and $\lnot A \lor B$, then in this case, $\lnot A$ is $\lnot\text{False}$, i.e. $\lnot A$ is a true statement. Since $\lnot A$ is a true statement, then so is $\lnot A \lor \text{True}$, and so is $\lnot A \lor \text{False}$ (if $B$ happened to be false), since a disjunctive ("or") statement evaluates to true whenever (at least) one of its disjuncts is true. Since $\lnot A$ is true, then $\lnot A \lor X$ is true no matter what the truth value of the statement $X$ happens to be.

A more "non-mathematician's" explanation might use a down-to-earth, not far-fetched example, like the following. Suppose the forecast predicts a 50/50 chance of rain tomorrow. I immediately assert that "If it rains tomorrow, then I'll take an umbrella with me to campus." Okay. So, now it's tomorrow:

scenario 1. It isn't raining, (and it doesn't rain at all), but I took my umbrella with me in case of rain. False --> True.

scenario 2. It isn't raining (and it doesn't rain at all). Good for me because I decided not to take my umbrella (since I have enough to lug around with me). False --> False.

Does that help?

share|improve this answer

In Aristotelian logic, we assume that every proposition is either true or false. This matches up poorly with common sense. E.g., it's not necessarily true or false that George W. Bush was the worst president in the history of the U.S.; it's a matter of opinion. However, Aristotelian logic is how we normally do mathematics. It's possible to do math using nonaristotelian logic, but you have to use different rules. A relatively readable and nontechnical book on this kind of thing is Graham Priest, An Introduction to Non-Classical Logic: From If to Is.

In Aristotelian logic, you can do proof by contradiction. If you can reason from not-P to a contradiction, then P is proved to be true. Therefore if you're working in Aristotelian logic and your formal system is not self-consistent, then your formal system proves all propositions to be true. That is, in Aristotelian logic, you can't have a little contradiction that only messes up things in one corner of the world. Any contradiction blows up your whole logical universe.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.