# Absurd = Inconsistent?

I think too much about the foundations of mathematics.

I'm not sure whether this is standard terminology, but I will refer, for some theory $\Gamma$, to statements as follows:

• A statement $P$ is inconsistent relative to $\Gamma$ if $P\ \&\ \neg P$ holds.
• A statement $P$ is absurd relative to $\Gamma$ if $\neg P\ \&\ \neg(\neg P)$ holds.

Now, clearly in terms of classical logic, these are the same, but it seems to me that this is not always the case. Consider, for instance, a statement $x \lt y\ \&\ x \ge y$. This I would consider an inconsistency, as even though both $P$ ($x \lt y$) and $\neg P$ ($x \ge y$) are sensible statements, we have a true contradiction.

It seems to me that this is a slightly concept notion than, say, Russel's Paradox. Denote the Russel Set $R = \{X | X \notin X\}$ and construct the predicate $P := R \in R$. The difference, as far as I can tell, is that neither $P$ nor $\neg P$ can be true, rather than both $P$ and $\neg P$ being true, which would be the case of an inconsistent statement. Thus I would term Russel's Paradox "absurd", but not "inconsistent".

This logic, of course, fails in classical logic, but what about other logics? Do absurdities (as defined above) necessarily reduce to inconsistencies in all logical systems? If not, suppose we admit absurdities into mathematics? If we view mathematics as a language (which, in many ways, it is), then it makes sense that we should be able to construct nonsensical sentences; it is not unthinkable that a "proof of absurdity" would even be possible in such a setting.

Lastly, suppose we admit the notion of absurdities, but not inconsistencies. Does naive set theory produce any true inconsistencies or does it produce only absurdities? In such a framework, we would have to augment proofs by contradiction with proofs that the statements under consideration are not absurd. Is this reasonable? What other sacrifices would need to be made?

Of course, most of my later questions are moot if my fundamental thesis that the absurd and the inconsistent can be different is false.

• I don't much like your terminology: "holds" where? In what theory? Perhaps better would be to say that "statement $P$ is inconsistent with $\Gamma$ if $\Gamma,P\models \neg P$"; and "statement $P$ is absurd relative to $\Gamma$ if $\Gamma,\neg P\models\neg\neg P$", or something like that. In any case, I think that in logic without excluded middle, you should find that inconsistent should imply absurd (though the converse may not be true). – Arturo Magidin Jan 23 '12 at 16:41
• I would say that the dichotomy you're looking for is between incompleteness and inconsistency (of a logical system). A system is incomplete if there is a sentence $P$ such that neither $P$ nor $\neg{P}$ can be proven. A system is inconsistent if there is a sentence $P$ such that both $P$ and $\neg{P}$ can be proven. Godel showed that any sufficiently powerful logical system is either incomplete or inconsistent; it seems likely that, say, Peano arithmetic and ZFC are incomplete rather than inconsistent. – mjqxxxx Jan 23 '12 at 16:45
• Arturo, good point...A little precision never hurts. I took your recommendation and added language specifying "relative to some theory $\Gamma$. – Dan M. Katz Jan 23 '12 at 16:46
• @DanM.Katz: Intuitionistic logic eschews the law of excluded middle; one way to think about is that it does not accept $\neg\neg P\to P$ as an axiom. Another is to think of $\neg$ as meaning "it is absurd that..." (as opposed to "it is false that") and so the argument is that "it is absurd that ($P$ is absurd)" is not the same thing as "$P$ is true". Contrary to popular misconception, intuitionistic logic does accept proof by contradiction, it just does not accept that double negations can be eliminated (and it also does not accept proofs of existence by contradiction). – Arturo Magidin Jan 23 '12 at 18:10
• You might consider defining "absurd" differently. One might define a statement as absurd relative to Γ iff (P & ¬P) does not always have truth value of false (I'm NOT using "not" here to mean logical negation), with inconsistent relative to Γ iff (P & ¬P) is true in the classical sense of true. There do exist plenty of non-classical logics where there does not exist any "inconsistency" but there do exist plenty of "absurdities" under this definition. – Doug Spoonwood Jan 23 '12 at 20:55

Suppose that $\Gamma \vdash \neg\varphi$ and that $\Gamma \vdash \neg\neg\varphi$. Since $\varphi$ is a wff, let $\psi$ be the wff $\neg\varphi$. Trivially, $\Gamma \vdash \psi \wedge \neg\psi$. So $\Gamma$ is inconsistent, and inconsistency and your 'absurd' coincide. This holds intuitionistically as well as classically: we don't need double-negation elimination to derive it.

Russell's paradox is a perfectly standard inconsistency: it allows us to derive a statement of the form $\varphi \wedge \neg\varphi$, but the law of noncontradiction states that $\neg (\varphi \wedge \neg\varphi)$, which is just a formal way of putting your statement that "neither $P$ nor $\neg P$ can be true".

(A minor terminological note: you talk about constructing "the predicate $P := R \in R$". This is not a predicate, but an atomic formula. Predicates must be predicable of something!)

• Wow, not sure how I missed that...I blame Mondays. Thanks! – Dan M. Katz Jan 23 '12 at 17:04
• In regards to the "predicate" remark: I was thinking about that as I wrote it, but I wasn't sure. Thanks for correcting me! :-) – Dan M. Katz Jan 23 '12 at 21:47

It's convenient to treat $\lnot P$ as an abbreviation for $P \to \bot$, where $\bot$ is a special formula that is false in every model. This is, in fact, how $\lnot P$ is defined in many intuitionistic systems.

Then $P \land \lnot P$ is the same as $P \land (P \to \bot)$ which implies $\bot$ by modus ponens. Also $\lnot P \land \lnot \lnot P$ implies $\bot$ for the same reason. Thus no model can satisfy either of these statements.

From this point of view, it becomes more clear why $P \to \lnot \lnot P$ is valid intuitionistically, because it corresponds to $P \to (P \to \bot) \to \bot$, which is yet another consequence of modus ponens. On the other hand $\lnot \lnot P \to P$ would correspond to $((P \to \bot) \to \bot )\to P$. This is not a simple consequence of modus ponens. In classical logic, we can prove it by reasoning about the truth value of $P$ being either True or False. That kind of reasoning by cases is unavailable intuitionistic logic, and in fact $\lnot\lnot P \to P$ is not provable there.

• In the last paragraph you have a sequence of $\rightarrow$'s which is unclear to me in its current form. Could you please add relevant parenthesis? – Asaf Karagila Jan 23 '12 at 19:02
• @Asaf: By convention $A \to B \to C$ means $A \to (B \to C)$. – Carl Mummert Jan 24 '12 at 1:46
• I was not aware of this convention before. I'm sure I'll be needing this information in the further future too, thanks! – Asaf Karagila Jan 25 '12 at 0:55