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to clarify the difference between this question and the supposed duplicate, absolutely nowhere do I mention the proof being attempted in the other question. It merely serves as an example related to this question.

Now, the way I view proof by contradiction is like this:

We have two logical statements $a$ and $b$ and we claim that "if $a$ then $b$".

We assume $b$ is false (given $a$) and we attempt to find some kind of contradiction with something we assume is true at the beginning (generally it tends to be proving $a$ is false).

Now, proof by contradiction can "fail" by resulting in no contradiction (i.e. that $a$ ends up being true when $b$ is false). However, this seems like a proof that the original statement is false. What I'm asking is more like how one detects or proves that when $b$ is false the truthfulness of $a$ is unprovable. Basically, that $a$ and $b$ are incapable of proving each other.

A good example of this is the parallel postulate of geometry. From what I heard (paraphrasing) "a failed proof by contradiction showed that there existed consistent systems where it was false and that it was unprovable from the other postulates". I'm not quite sure how that worked. Could someone please explain (in general) how one proves a proof by contradiction fails this brutally? I'm not asking specifically how it was done for the parallel postulate, but just how one goes about doing it in general?

user64742
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2 Answers2

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The sentence you paraphrased isn't true on the nose - it's describing an important phenomenon though.

Let's say you're trying to prove "$A$" by contradiction; so you imagine that $A$ were false, and try to derive an absurdity from that. You do this for a while, write a few pages, and get nowhere. Your attempt has failed.

. . . But you don't know that it has to fail! Maybe if you just kept going for a page, or two, or two hundred, you'd find a way to get that contradiction you're looking for! So how do you know it can't work?

Well, this is where consistency comes in. To show that you can't prove $A$ - by contradiction or otherwise - it's enough to show that there's a model of your axioms, together with $\neg A$. For instance, in the case of geometry, the way we know that you can't prove the parallel postulate from the other four is that we know that there are models of the other four postulates + the negation of the parallel postulate (e.g. hyperbolic geometry). This is a standard technique in logic.

Now, building a model of $\neg A$ doesn't really have anything to do with a failed proof attempt of $A$ directly - but in practice, the one tends to lead to the other! Namely, if I'm trying to derive a contradiction from $\neg A$, what I do is try to picture in my head what things would be like if $A$ were false. Sometimes this leads me to a contradiction; if it doesn't, though, this can lead me to an understanding of what a model of $\neg A$ might look like! And this in turn can help me figure out how to build a model of $\neg A$.

This is what is meant by the statement you give. People tried to prove the parallel postulate by contradiction. They didn't succeed, and in fact in the course of their attempt figured out how to build a model where the parallel postulate fails. The existence of this model proves that their original attempt was doomed.

(Incidentally, I have also heard this statement about the parallel postulate, but I'm not sure if it's really historically accurate; I suspect it's a bit of a simplification. But, this is what that statement means.)

Note that there's nothing special about proof by contradiction here - anytime I try to prove $A$ and fail, I might gain some intuition of how to build a model of $\neg A$ (which wouldn't tell me that $A$ is false, just that $A$ is not provable from the axioms I've listed so far). More generally, failed attempts to prove one thing often lead to a successful attempt to prove a very different thing - consider e.g. Kleene's proof of the recursion theorem, which emerged following a failed attempt to build an intuitively-computable, non-general-recursive function.

Noah Schweber
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  • Interestingly, it turns out that the above idea - show that $A$ isn't provable by constructing a model of $\neg A$ - is universal in a precise sense: if $A$ is not provable in some theory $T$, then there is a model of $T+\neg A$. Conversely, if $A$ is provable in $T$, then there are no models of $T+\neg A$. These two facts are the completeness and soundness theorems, respectively. – Noah Schweber Jan 21 '17 at 01:08
  • +1 Your longer answer than mine addresses the philosophy of discovery as well as the existence of the useful model. – Ethan Bolker Jan 21 '17 at 01:37
  • @TheGreatDuck: Just a note, those two facts that Noah mentioned are for classical first-order logic. There may be corresponding theorems for other logics (like intuitionistic logic with Kripke semantics), or there may not. – user21820 Jan 30 '17 at 05:06
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Here is

how one proves a proof by contradiction fails this brutally

in the context of the parallel postulate.

All attempts to show that asserting the falsity of the parallel postulate leads to a contradiction are doomed to fail because there is a perfectly good geometry in which the parallel postulate happens to be false while all Euclid's other axioms are true. You can build a model of that geometry inside Euclidean geometry, so any contradiction could happen only because the other axioms of Euclidean geometry are inconsistent.

This is in fact what Lobachevsky (and Gauss) discovered. One of the nicest models is the Poincare disk model.

You asked for a general method. This argument is an example. There can be no proof by contradiction that proposition $P$ is false if you can find a situation where all your assumptions are true and $P$ is false.

Ethan Bolker
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