Absurd = Inconsistent?

Question

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.

Thanks in advance!

Answer 1

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!)

Answer 2

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.