# some question about logic that the premise is false.

Is it just for convenience that define false statements imply anything?

if yes, why it would be defined like this?

-
Suppose you go to the park and see a sign that says "All dogs must..." If you don't have a dog, do you have to read the rest of the sign? –  Ted Jan 8 '13 at 6:13
I'd say that false statements don't imply anything one way or the other. If $p$ is false, then $(p\rightarrow q)$ ("whenever $p$ is true, $q$ must be true") is satisfied whether or not $q$ is true. –  mjqxxxx Jan 8 '13 at 6:13
It is like an 'if' statement in a programming language. –  copper.hat Jan 8 '13 at 6:15
@mjqxxxx What you said is false (no pun intended). A false proposition implies any proposition. –  Vectk Jan 8 '13 at 6:20
@Brian: Actually, what he said is true, he just used 'imply' in a more normal-English sense :) –  Eric Stucky Jan 8 '13 at 6:27

Well, a big reason that we chose this convention is that we are sort of short on options:

• We could say that a false premise implies that the implication is true exactly when the conclusion is true, but that would be odd because then the premise doesn't do anything.
• We could say that a false premise implies the implication is true exactly when the conclusion is false, but eww.
• We could say that a false premise implies the implication is always true, which is what most people who think about the alternatives do.
• We could say that a false premise implies the implication is always false, but if we do this then $p\to q$ has the same truth table as $p\wedge q$, which isn't bad but it seems like there is something more that we want out of an implication than a simple 'and' statement.
• We could reject the law of the excluded middle so that the implication is neither true nor false. This turns out to be a valid option, but also eww. More practical reasons to dislike this is that it destroys double-negation $(\sim\sim\! p$ is $p)$ and therefore contrapositive and contradiction proofs.

So you have to pick one. They're a sorry lot, I admit, but we're stuck with them, and one has proven to be more realistic and pragmatic than the others.

-
there is another alternative: you can avoid an attempt to define the meaning of implication from falsehood and reject it as nonsensical. After all, who cares what the truth value of P-->Q is when P is false? (e.g., if you pay me I will give you an apple. Now, you did not pay me, so what the %&^$do you want?) – Ittay Weiss Jan 8 '13 at 8:21 @IttayWeiss: Is that substantially different than rejecting the law of the excluded middle (by creating a third truth value and assigning it to the implication)? I don't think so, but I admit that I haven't studied such things :/ – Eric Stucky Jan 8 '13 at 16:50 The law of excluded middle can fail in, at least, one of three ways. 1) Introduce new truth values 2) declare that P does not have to have any truth value at all 3) allow P to be both true and false at the same time. Now, introducing new truth values requires explaining what exactly these things are (it's not quite true, nor false, it sort of in between.....). Sticking to just two truth values thus has merits. Options 2 and 3 then allow simple solutions to problems. Consider Russele's paradox, P. What if P is simply true and false at the same time without it have any ill consequence? – Ittay Weiss Jan 9 '13 at 0:08 Quine asserts a similar idea in his “Methods of Logic”: “An inconsistent schema implies every schema and is implied by inconsistent ones only.” This follows naturally from the manner in which he defines implication. Specifically he says, “implication is validity of the conditional,” where validity was previously defined as being true under all interpetations. With this definition of implication, a schema of the form ‘${\bot} \to p$’ is valid because it is true whether$p$is interpreted true or false, therefore ‘$\bot$’ implies ‘$p$’. - It is not entirely clear from your question what you mean so I'm adding this answer too, just for the sake of completeness. I think you might be referring to the fact that in classical logic a contradiction implies any sentence. This is not the same as saying that a false statement implies anything. That is actually not true. If$P$is false then$P\implies Q$is a true statement but one can't conclude that$Q$is true. If, however,$P$is both true and false (i.e., is a contradiction) then since$P\implies Q$is true one may now conclude by Modes Ponens that$Q\$ is true, and thus a contradiction proves any statement. This is called the explosion principle.

There are several reasons to feel somewhat not at ease with this principle and there are ways to exclude this principle and retain a very useful logical system that is tolerant to contradictions, without rendering the entire system useless. Such logics are called paraconsistent. A nice expository article on paraconsistency can be found here: http://plus.maths.org/content/not-carrot

-

You got it backward: a proposition does not imply everything because it is false, but a proposition is false because it implies everything. We want a formal system to single out a strict subset of valid propositions. A proposition that implies every other proposition cannot be in this subset, as it would make every other formula valid too. Therefore, those propositions that imply everything, are false.

-