Why in an inconsistent axiom system every statement is true? (For Dummies)

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.

Solutions Collecting From Web of "Why in an inconsistent axiom system every statement is true? (For Dummies)"

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!

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.

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.

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.

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.

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.

‎”inconsistent” means that if you ask the same question twice, you get a 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 purpose 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 its negation, we cannot do it within the accepted meaning of the definitions “consistent” and “negation”