When $p$ and $q$ are propositions such that "if $p$ then $q$" (which is an implication) is false only in the condition when $p$ (which is hypothesis) is true but $q$ (which is conclusion) is false.

But I couldn't understand why will the implication be true when the hypothesis itself is false. Shouldn't the implication be false when the initial assumption or hypothesis is false ?

  • $\begingroup$ You are wrong. The implication is false only when $p$ is true and $q$ is fase. $\endgroup$ Aug 16, 2017 at 11:41
  • $\begingroup$ Sorry for the typo. I was trying to say why would implication be true when hypothesis is false or in short why would implication be false only when p is true but q is falsw $\endgroup$ Aug 16, 2017 at 11:52
  • $\begingroup$ @Piquito I think the proposition "All triangles are Rectangles" is false because no triangle can be a rectangle and no rectangle can be a triangle. $\endgroup$ Aug 16, 2017 at 11:57
  • $\begingroup$ What about ALL TRIANGLES ARE RECTANGULAR $\Rightarrow$ SOME TRIANGLES ARE RECTANGULAR? $\endgroup$
    – Piquito
    Aug 16, 2017 at 12:03
  • $\begingroup$ The closure of this question happened when I was typing an answer... Shoot :). Note that, if you let "$A \Rightarrow B$ holds" be defined by "$\overline{A}$ or $B$ holds", then from the definition of "or" you can answer your question directly. Intuitively speaking, you may view the rule of inference this way. If your teacher promise you (weirdly) that "if you get an A+ this time, then you can graduate directly from then on under my permission", then when will he break the promise? When and only when you get an A+ that time and he does not make your graduation happen! $\endgroup$
    – Yes
    Aug 16, 2017 at 12:24

2 Answers 2


This has been answered many many times on this site, but here is a simple reason. If one would use any other truth table for $P\to Q$ it would no longer mean what it is supposed to. If the hypothesis $P$ is true, then everyone agrees that $P\to Q$ should hold if and only if $Q$ does. Now suppose the hypothesis $P$ is false. If one would define (as you suggest) $P\to Q$ to be false in this case, regardless of $Q$, then $P\to Q$ would be true only if both $P$ and $Q$ are true, making it a symmetric operation for which we already have a name (namely "and", written $P\land Q$). If instead when $P$ is false we would define $P\to Q$ to be true only if $Q$ is, then the meaning of $P\to Q$ would ignore the status of the hypothesis $P$ completely, and be equivalent to $Q$. Finally if when $P$ is false we would define $P\to Q$ to be true only if $Q$ is false, then $P\to Q$ would be true if any only if $P$ and $Q$ have the same truth value, again a symmetric relation that does not reflect what we mean by implication. So the only reasonable option is the define, when the hypothesis is false, the relation $P\to Q$ to be true regardless of $Q$.

And this actually captures what one wants to say pretty well: if the hypothesis is true then we claim the conclusion must be true as well, but if the hypothesis is false, we did not mean to claim anything, so we shouldn't be called out for making a false claim in such cases.


The implication is the statement "if p then q". It's true if "every time" p is true, q is also true. Since p is "never" true, it satisfies the statement, so the implication is true. I wrote "every time" and "never" because the value of p and q never changes per se.

It can be also demonstrated with the empty set. We can say that all numbers in the empty set are even. Otherwise it wouldn't be a subset of the set of even numbers. But it is. So this can be reformulated like "if the empty set contains a number A, then A is even." p = "the empty set contains a number A" q = "A is even" So the implication must be true, or the empty set won't be a subset of even numbers.


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