# The Universe of Discourse

Tue, 01 Aug 2023

Lately I have been thinking about the formula

$$((P\to Q)\land (\lnot P \to Q)) \to Q \tag{\color{darkgreen}{\heartsuit}}$$

which is a theorem of classical logic, but not of intuitionistic logic. This shouldn't be surprising. In CL you know that one of !!P!! and !!\lnot P!! is true (although perhaps not which), and whichever it is, it implies !!Q!!. In IL you don't know that one of !!P!! and !!\lnot P!! is provable, so you can't conclude anything.

Except you almost can. There is a family of transformations !!T!! where, if !!C!! is classically valid !!T(C)!! is intuitionistically valid even if !!C!! itself isn't.

For example, if !!C!! is classically valid, then !!\lnot\lnot C!! is intuitionistically valid whether or not !!C!! is. IL won't prove that !!(\color{darkgreen}{\heartsuit})!! is true, but it will prove that it isn't false.

I woke up in the middle of the night last month with the idea that even though I can't prove !!(\color{darkgreen}{\heartsuit})!!, I should be able to prove !!(\color{darkred}{\heartsuit})!!:

$$((P\to Q)\land (\lnot P \to Q)) \to \color{darkred}{\lnot\lnot Q} \tag{\color{darkred}{\heartsuit}}$$

This is correct; !!(\color{darkred}{\heartsuit})!! is intuitionistically valid. Understanding !!\lnot X!! as an abbreviation for !!X\to\bot!! (as is usual in IL), and assuming $$\begin{array}{rlc} P\to Q & & (1) \\ \lnot P\to Q & & (2) \\ \lnot Q & (≡ Q\to\bot) & (3) \end{array}$$

we can combine !!P\to Q!! and !!Q\to\bot!! to get !!P\to\bot!! which is the definition of !!\lnot P!!. Then detach !!Q!! from !!(2)!!. Then from !!Q!! and !!(3)!! we get !!\bot!!, and discharging the three assumptions we conclude:

\begin{align} \color{darkblue}{(P\to Q)\to (\lnot P \to Q)} & \to \color{darkgreen}{\lnot Q \to \bot} \\ ≡ \color{darkblue}{((P\to Q)\land (\lnot P \to Q))} & \to \color{darkgreen}{\lnot\lnot Q} \tag{\color{darkred}{\heartsuit}} \end{align}

But what is going on here? It makes sense to me that !!(P\to Q)\land (\lnot P \to Q)!! doesn't prove !!Q!!. That wasn't the part that puzzled me. The puzzling part was that it would prove anything more than zero. And if !!(P\to Q)\land (\lnot P\to Q)!! can prove !!\lnot\lnot Q!!, then why can't it prove anything else?

This isn't a question about the formal logical system. It's a question about the deeper meaning: how are we to understand this? Does it make sense?

I think the answer is that !!Q\to\bot!! is an extremely strong assumption, in fact the strongest possible statement you can make about !!Q!!. So it's easist possible thing you can disprove about !!Q!!. Even though !!(P\to Q)\land(\lnot P\to Q)!! is not enough to prove anything positive, it is enough, just barely, to disprove the strongest possible statement about !!Q!!.

When you assume !!Q\to \bot!!, you are restricting your attention to a possible world where !!Q!! is actually false. When you find yourself in such a world, you discover that both !!P\to Q!! and !!\lnot P\to Q!! are much stronger than you suspected.

My high school friends and I used to joke about “very strong theorems”: “I'm trying to prove that a product of Lindelöf spaces is also a Lindelöf space” one of us would say, and someone would reply “I think that is a very strong theorem,” meaning, facetiously or perhaps sarcastically, that it was false. But facetious or sarcastic, it's funny because it's correct. False theorems are really strong, that's why they are so hard to prove! We've been trying for thousands of years to prove a false theorem, but every time we think we have done it, there turns out to be a mistake in the proof.

My puzzlement about why !!(P\to Q)\land (\lnot P\to Q)!! can prove anything, translated into computational language, looks like this: I have a function !!P\to Q!! (but I don't have any !!P!!) and a function !!\lnot P\to Q!! (but I don't have any !!\lnot P!!). The intutionistic logic says that I can't use these functions to to actually get any !!Q!!, which is not at all surprising, because I don't have anything to use as arguments. But IL says that I can get !!\lnot\lnot Q!!. The question is, how can I get anything from these functions when I don't have anything to use as arguments?

Translating the proof of the theorem into computations, the answer one gets is quite unsatisfying. The proof observes that if I also had a !!Q\to\bot!! function, I could compose it with the first function to make a !!P\to\bot\equiv \lnot P!! which I could then feed to the second function and get !!Q!! from nowhere. Which is very strange, since operationally, where does that !!Q!! actually come from? It's manufactured by the !!\lnot P\to Q!! function, which was rather suspicious to begin with. What does such a function actually look like? What functions of this type can actually be implemented? It all seems rather unlikely: how on earth would you turn a !!P \to \bot!! value into a !!Q!! value?

One reasonable answer is that if !!Q = \lnot P!!, then it's easy to write that suspicious !!\lnot P\to Q!! function. But if !!Q=\lnot P!! then the claim that I also have a !!P\to Q!! function looks extremely dubious.

An answer that looks good at first but flops is that if !!Q=\mathtt{int}!! or something, then it's quite easy to produce the required functions, both !!P\to Q!! and !!\lnot P\to Q!!. The constant function that always returns !!23!! will do for either or both. But this approach does not answer the question, because in such a case we can deduce not only !!\lnot\lnot Q!! but !!Q!! itself (the !!23!! again), so we didn't need the functions in the first place.

Is the whole thing just trivial because there is no interesting way to instantiate data objects with the right types? Or is there some real computational content here? And if there is, what is it, and how does that translate into the logic? Does this argument ever allow us to conclude something actually interesting? Or is it always just reasoning about vacuities?

### Note

As far as I know the formula !!(\color{darkgreen}{\heartsuit})!! was first referred to as “Gantō's Axe” by Douglas Hofstadter. This is a facetious reference to a certain Zen koan, which says, in part:

Ganto picked up an axe and went to the hut where the two monks were meditating. He raised the axe, saying, “If you say a word I will cut off your heads. If you do not say anything, I will also behead you.”

(See Kubose, Gyomay M. Zen Koans, p.178.)

$$((P\to Q)\land ((P\to R) \to Q)) \to {(Q\to R)\to R}$$