TITLE: Looking for a simple proof of the independence of the law of excluded middle
QUESTION [7 upvotes]: I have seen a number of excellent posts on the difference between intuitionist propositional logic (IPL) and classical propositional logic (CPL), all of which state that IPL is agnostic on the law of excluded middle (LEM) or its equivalent forms. I have not seen a clear, simple demonstration that LEM is independent of IPL. Can someone point me in the right direction? It looks like maybe one way to do this involves the BHK interpretation of IPL, with some results about non-computability.
Can someone point me toward a place where that argument is spelled out more? Is there a simple alternative to it? Thanks!

REPLY [11 votes]: The others answers give simple semantic proofs showing that LEM is independent of intuitionistic logic.
However, you don't always need to construct a model to show non-derivability. There are simple, clear proof-theoretic arguments showing the same thing, which have the advantage that they work for intuitionistic first-order predicate logic as well, without any additional work or need for modification.
One of them relies on the following important property satisfied by the intuitionistic calculus.
Disjunction property: For any formulae $M,N$, if there is an intuitionistic derivation of $\vdash M \vee N$, then we can find a derivation of either $\vdash M$ or $\vdash N$.
By the disjunction property, if $\vdash A \vee \neg A$ was provable, then we would be able to prove either $\vdash A$ or $\vdash \neg A$. But it's clear that neither of these have (even classical) derivations. Hence, $\vdash A \vee \neg A$ is not provable.
Okay, that's quick and simple, but how do we prove the disjunction property? Well, intuitionistic logic is usually defined by means of either a sequent calculus deduction system, a natural deduction system, or a Hilbert system. For the sequent calculus, we can prove it simply by looking at cut-free derivations (see [1] for a more general version). For natural deduction, we can either reduce it to the sequent calculus, or prove it directly by proof normalization (see [2]). The Hilbert system is also reducible to the sequent calculus. So I'll give the short sequent calculus version:
Recall that the sequent calculus has the following two rules for introducing $\vee$ on the right of the turnstile $\vdash$:
$$\frac{\Gamma \vdash A}{\Gamma \vdash A \vee B}\vee\!R_1$$
$$\frac{\Gamma \vdash B}{\Gamma \vdash A \vee B}\vee\!R_2$$
If $\vdash M \vee N$ has a derivation, then by the cut-elimination theorem it has a derivation that doesn't use the cut rule. Look at the last rule of a cut-free derivation of $\vdash M \vee N$. It cannot be cut, so it can only be $\vee\!R_1$ or $\vee\!R_2$. If it's $\vee\!R_1$, then the previous line was $\vdash M$, so $\vdash M$ has a derivation as claimed. Otherwise, the last rule is $\vee\!R_2$ and the previous line was $\vdash N$, so $\vdash N$ has a derivation. This proves the disjunction property.
edit: As commenters pointed out, it's worth mentioning that the negation of LEM, $\vdash \neg (A \vee \neg A)$, cannot be derived in intuitionistic logic either. To show this, all you have to do is notice that $\neg (A \vee \neg A) \vdash A$ is provable in intuitionistic logic. So if $\vdash \neg (A \vee \neg A)$ was provable as well, you'd be able to use the cut rule to conclude that $\vdash A $ would also be provable. But as we observed above, during the proof of the disjunction property, $\vdash A$ is not derivable. Thus, $\vdash \neg (A \vee \neg A)$ is not derivable either.
[1] A. S. Troelstra, H. Schwichtenberg: Basic Proof Theory, 2nd edition, 2000. ISBN 9781139168717
[2] D. Prawitz: Natural Deduction: A Proof-Theoretical Study, Dover Publications, 1965. ISBN 9780486446554x