What is a constructive proof of $\lnot\lnot(P\vee\lnot P)$?
Glivenko's theorem says that $\lnot\lnot P$ is a theorem of intuitionistic logic whenever $P$ is a theorem of classical logic. Is it closely related to the so-called Gödel–Gentzen negative translation which embeds classical logic into intuitionistic logic.
Since $P\vee\lnot P$ is a well-known theorem of classical logic, I expect that by Glivenko's theorem, $\lnot\lnot(P\vee\lnot P)$ is provable in intuitionistic logic. But I cannot find a proof! I must be overlooking something obvious.
If $\lnot\lnot(P\vee\lnot P)$ is indeed provable in IL, I would like to see a natural deduction or sequent calculus proof of it, or especially a construction of a typed $\lambda$-calculus term with the type $((P\vee(P\to \bot))\to\bot)\to\bot$.
If it is not provable in IL, what have I misunderstood?
Addendum: I cross-checked the proposition with the usual model, letting $P$ be a subset of $\mathbb R$, and interpreting $\vee$ as set union and $\lnot x$ as the interior of the complement of $x$, and the proposition $\lnot\lnot(P\vee\lnot P)$ did seem to come out as all of $\mathbb R$ for any choice of $P$, so if I have made a mistake here too I would like to know what subset of $\mathbb R$ is the counterexample.
Per the BHK interpetation, a proof of $(P \lor (P \to \bot)) \to \bot$ is pair of procedures, one for each disjunct in the hypothesis.
The first, $s_1$, is a procedure which takes a proof of $P$ as input and produces a proof of $\bot$. In other words, the first is a proof of $P \to \bot$.
The second, $s_2$, takes any proof of $P \to \bot$ as input and produces a proof of $\bot$. In other words, it is a proof of $(P \to \bot) \to \bot$.
By composing these to form $s_2(s_1)$, we obtain a procedure which takes no input and produces a proof of $\bot$.
The procedure just described (composition) takes any proof of $(P \lor (P \to \bot)) \to \bot$ and produces a proof of $\bot$. This means that $((P \lor (P \to \bot))\to \bot) \to \bot$ is provable.
This description shows how to write a $\lambda$ expression of the appropriate type. All that it does is to break the input into its two components and run the second one with the first as input. In particular, if we view $(P \lor \lnot P) \to \bot$ as a type $p(n,s)$ such that $\lambda s.p(1,s)$ is of type $P \to \bot$ and $\lambda\,s.p(2,s)$ is of type $\lnot P \to \bot$ then $$\lambda\, p(n,s) . p(2,\lambda\, s.p(1,s))$$ is of type $((P \lor \lnot P) \to \bot) \to \bot$
Here's a natural deduction proof:
- $\lnot (P \lor \lnot P)$ [Assumption]
- $\lnot P \land \lnot \lnot P$ [De Morgan]
- $\lnot P$ [$\land$ elimination]
- $\lnot \lnot P$ [$\land$ elimination]
- $\bot$ [$\bot$ introduction from 3., 4.]
- $\lnot \lnot (P \lor \lnot P)$ [$\lnot$ introduction]