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I would like to know if someone can explain in a somehow down to earth (almost logic free) way why is it true that in an axiom system where there is some statement $P$ such that $P$ and its negation $\lnot P$ are true, then every statement in the system is true?

I'm not sure if this can be done, but basically since I don't know any formal logic at all, I'm interested in seeing if at least the argument can be conveyed in an intuitive way, or if the idea can be explained without talking about first or second order logic and using symbols like $\top$, $\bot$, and $\vdash$.

This previous question is like the formal version (which I don't understand) so maybe my question can be thought of as a version for dummies of that question.

Thanks a lot in advance.

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It really comes down to the implication. If you accept that "false implies false" and "false implies true" are both true, then the result follows from basic modus ponens. The problem is "material implication" does not match well with linguistic implication, which makes a "for dummies" or "intuitive" explanation somewhat difficult. Much like the fact that "a or b" can be true when both a and b are true, which strikes a lot of people as very strange (being used to exclusive or from natural language). –  Arturo Magidin Apr 2 '11 at 2:39
    
Please feel free to edit the tags since I didn't know which tags to put. Maybe an "elementary logic" tag would have been more appropriate. And if someone feels the expression "for dummies" I used in the title makes it seem impolite, I would gladly remove it from the title. –  Adrián Barquero Apr 2 '11 at 2:40
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One comment: this might just be a quibble or it might actually be helpful: formal systems don't in themselves deal with truth: rather they deal with provability. It sounds really weird to say that $P$ and $\lnot P$ are both true in your formal system. Rather you are saying that $P$ and $\lnot P$ are both provable, which is not contradictory but even rather banal...after all, maybe your formal system just proves everything(!). The correspondence between truth and provability is addressed in formal logic -- notably, Godel's Completeness Theorem -- but that's another story entirely. –  Pete L. Clark May 20 '11 at 9:07

7 Answers 7

up vote 46 down vote accepted

I'll try to say this all in plain English:

Let's say we decide to accept the following two facts: (1) "I am a fish", and (2) "I am not a fish". Just keep those in mind.

Now let's pick any old statement, say: (3) "You can fly". Now let's prove that the statement is true!

Alright, we've already accepted that (1) "I am a fish". Of course, any time I have a true statement P, I can make a new true statement by making the statement "P or Q is true." Because to check if an 'or' statement is true, I only need to check that one of them is true. (If I tell you "My name is Dylan OR I can spit fire," you don't need to wait around with a fire extinguisher to tell if that statement is true. It's true because the first part of it is true).

So by this logic, the statement (4) "I am a fish or you can fly" must be true (since the first part is true.)

OK, but now let's say, in general, I have some 'or' statement "P or Q" and I know for a fact that the whole statement is true. If I also know that P is false then I can conclude that Q is true. Right? Because an 'or' statement is true if and only if at least one of the statements inside it is true, so if I rule out one of them the other one must be true. (So if I always tell the truth and I tell you that you have a billion dollars in your bank account OR I just ate a sandwich, you can check your bank account and quickly conclude that I just ate lunch... unless you're very wealthy.)

Alright, so far so good. We know the statement "I am a fish or you can fly" is definitely true. But wait, we also know that the statement "I am a fish" is false (remember, it's one of the things we assumed in the very beginning!). So that means, by what we just talked about, that the statement "You can fly" must be true.

So voilà! Using the magic of a contradictory system, we've proven you can fly!

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Thanks a lot Dylan. This is a really nice answer. Besides you made me laugh a couple of times while reading it. It's a shame I can't upvote twice =) –  Adrián Barquero Apr 2 '11 at 4:15
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This is great! I got it, so it must be simple enough for anybody to understand. –  Uticensis Apr 2 '11 at 4:25
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+1 for for the reference to dylan spitting hot fire...how you gone choke a legend?! –  Jason Apr 2 '11 at 7:46
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YOu can probably find on-line an account of the time Bertrand Russel proved that he was the Pope. –  GEdgar May 20 '11 at 12:43
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Maybe one should elaborate on the possibly subtle difference between "I am not a fish" being true and "I am a fish" being false ... –  Hagen von Eitzen Sep 2 '12 at 11:55

It is somewhat misleading to say "every statement is true" about an inconsistent theory. This might be a point of confusion in this question, and in general the difference between "truth" and "provability" causes many other confusions, so we have to be careful to distinguish them.

"Truth" is a property that a statement has in a particular model. In other words a particular statement is either true or false in a particular model, assuming the statement is written in a formal language compatible with the model.

An inconsistent theory has no models at all. If you have no models, there's no model in which any particular statement can be true. It is correct to say that every statement in the language of the theory is provable from an inconsistent theory, and that every statement in the language of the theory is semantically entailed by the theory. But it's an abuse of language to say that every statement is "true" in the theory: true in what model?

Sometimes, when people are writing an informal proof, they say things like "assume $A$ is true" or "assume $B$ is false". But these are just figures of speech; the actual proof system usually has other ways of dealing with hypotheses than to mark them as "true" and "false". Alternatively, you can view those sayings as abbreviated forms of "assume $A$ is true in some fixed, unspecified model", "assume $B$ is false in our fixed, unspecified model". That interpretation of the informal proof is fine for any consistent theory. But that interpretation is more difficult for inconsistent theories. Because there are no models, it will be a counterfactual statement.

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The answers in the prior question all employ syntactic consequence, i.e. consequence by proof in some formal system. It may be clearer to instead consider this as a semantic consequence, i.e. $\rm\ S\models Q\ $ means that $\rm\:Q\:$ is true in every model ("possible world") where every member of $\rm\:S\:$ is true. Thus $\rm\ \{P, \lnot P\}\models Q\ $ is true vacuously, since there are no models where both $\rm\:P\:$ and $\rm\:\lnot P\:$ are both true, i.e. there is no model witnessing a counterexample to that consequence, where every element of $\rm\:S\:$ is true but $\rm\:Q\:$ is false. You can find some further discussion in the Wikipedia article on this Principle of Explosion.

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Thanks a lot Bill. I guess I'll have to dig up some terminology from formal logic though. –  Adrián Barquero Apr 2 '11 at 3:30

In most formal logics, the rules of the system force the principle of explosion (as pointed out by Bill Dubuque). However, there are interesting logics where contradictions do not necessarily propagate like this. This kind of logic is called paraconsistent logic.

Your lack of intuition about this question is perhaps because people usually are able to deal gracefully with contradiction. In contrast, classical logic defines away contradiction: this makes the deductive system much tidier, but it makes it harder to use logic to model real-world situations.

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In basic sentence logic: there is a sentence $P$ such that $P$ implies $S$ and $P$ implies $\lnot S$. Then, start assuming any statement $\lnot Q$ in your system. Then introduce the sentence $P$, from which $S$ and $\lnot S$ follow. Then the assumption of $Q$ leads to the conclusion $S \land \lnot S$, so by reduction, $\lnot Q$ follows. You can do this for any sentence $Q$ in your system.

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In an inconsistent system you can prove a false proposition. Suppose $F$ is a false proposition. Let $P$ be a proposition. We will show that $P$ is true.

Since $F$ is false, $\lnot F$ is true. It follows that $\lnot F \vee P$ is true. Why? Because for every true propostion $X$, $X\vee Y$ is also true for any proposition $Y$.

What can we deduce from $\lnot F \vee P$? We assumed that $F$ is true, from which it follows that $\lnot F$ is false. But if $\lnot F \vee P$ is true, which we showed, at least one of $\lnot F$ or $P$ must be true. Since $\lnot F$ is false, $P$ must be true.

Alternative proof:

$X$ implies $Y$ is equivalent to $\lnot\ X \vee Y$. Let's say $X$ is false. It follows that $\lnot X$ is true, which implies $\lnot X \vee Y$, which is equivalent to $X$ implies $Y$. Since we assumed $X$ to be true, we can conclude that $Y$ is true.

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‎"inconsistent" means that if you ask the same question twice you get different answer. To say that you get the answer "true" every time, is "consistent". The definition of "negative proposition" (the opposite of a proposition) is meant for the pupose of getting a negative result in the event the proposition gets a positive. So if we want to make a system which is "consistent" and gives the same result for a proposition and it's negation, we cannot do it within the accepted meaning of the definitions "consistent" and "negation"

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I asked ZFC the following question "Is the continuum of size $\aleph_1$?" and once it said "Yes!" and once it said "No way". This answer is too informal that it loses the mathematical point to it. –  Asaf Karagila May 20 '11 at 7:33
    
@Asaf Karagila: I do not see why it is too - informal to have a mathematical point to it. It presents the intuition behind some of the concepts raised in the question. The person asking the question put the intuition tag so this more lay style suites this post. Formality is relavent to other tags. –  Vass May 20 '11 at 11:10
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What do you mean by "ask"? If you ask whether the statement is provable, you will get the same answer every time. If you ask whether it's true in a particular model, you'll get the same answer every time as well. What question is being asked twice but getting different answers? –  Carl Mummert May 20 '11 at 11:19
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So what then does it mean "Inconsistent Axiom System"? any examples? –  Panos May 21 '11 at 10:01
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Furthermore: To test a statement in a system and see if it is true or false, you suppose it is consistent. If you know it is "Inconsistent", what do you expect? –  Panos May 21 '11 at 10:04

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