TITLE: What's up with the Sheffer stroke axiom?
QUESTION [8 upvotes]: While reading some old paper on the foundations of set theory, I came across a symbol $\mid$ that I eventually determined was the Sheffer stroke, which is a fancy word for NAND.
Wikipedia, and also the paper, has only this to say about working with the Sheffer stroke:

All [well-formed formulas] of the form $$((U \mid (V \mid W)) \mid ((Y \mid (Y \mid Y)) \mid ((X \mid V) \mid ((U \mid X) \mid (U \mid X)))))$$ are axioms. Instances of $$(U \mid (V \mid W)),\, U \vdash W$$ are inference rules.

This is a remarkable achievement in concise formulation: it leaves out everything that I understand about axiomatic proofs. My question could be summarized as "what the heck?" but for the sake of people trying to answer it, I'll try to be a little bit more precise.
How do you actually work with the thing above? What do the parts of a proof in language I understand become when you use the Sheffer stroke instead? If the only axiom is the disaster above, how do you ever conclude anything that makes sense?
I'm guessing that an example of a proof of something, anything using this formal system, with an explanation, would answer my question and address my confusion.

REPLY [6 votes]: Doug Spoonwood's answer was very helpful, as was Jean Nicod's paper introducing the Sheffer stroke axiom, but I'm going to accept the former because Nicod doesn't appear interested in StackExchange karma.
I'm going to post my own answer describing what I learned about working with the Sheffer stroke and with this axiom in particular. My personal goal was to prove $Y \mid (Y\mid Y)$, since it seemed like a very simple fact that should be true; below, I prove that, and a few other things, in potentially human-readable form.
How I learned to stop worrying and love the Sheffer stroke
Some nicer notation, partly borrowed from Nicod:

$x \mid y/z$ as shorthand for $x \mid (y \mid z)$; in general, $/$ is a version of $\mid$ that is higher in order of operations, to avoid parentheses.
$\overline{x}$ as shorthand for $x \mid x$; this can be read as (and is equivalent to, outside the system) "not $x$". Note that from $x$ and $x \mid \overline y$, we can infer $y$.
$\pi_x$ as shorthand for $x \mid \overline x = x \mid x/x$. This should be true for all $x$.
$[x,y]$ as shorthand for $x/y \mid \overline{y/x}$. This should be true for all $x,y$; from $[x,y]$ and $x\mid y$, we can "swap $x$ and $y$" and infer $y \mid x$.
$[x,y,z]$ as shorthand for $(x\mid y/z) \mid \overline{z/y \mid x}$. This should be true for all $x,y,z$; from $[x,y,z]$ and $x \mid y/z$, we can "reflect $x,y,z$" and infer $z/y \mid x$.

(Nicod uses $\pi$ for a specific $\pi_t$; $[x,y]$ and $[x,y,z]$ are all mine, inspired by commutator brackets in a completely different area of math.)
In this answer, I'll prove $\pi_x$, $[x,y]$, and $[x,y,z]$ for all $x$,$y$, and $z$.
Lemma 1. $[x, \pi_y]$ for all $x,y$.
Proof. The single axiom we've got can be written as
$$
    (u \mid v/w) \mid \pi_y / ( x/v \mid \overline{u/x}).
$$
In particular, if we set $u=v$ in the axiom, we have
$$
    (u \mid u/w) \mid \pi_y / [x,u].
$$
If we can prove anything of the form $u \mid u/w$, we can deduce $[x,u]$. In particular, setting $u=v=w=x=y$ in the axiom gives us $\pi_y \mid \pi_y / [y,y]$, which has that form; therefore, we have $[x, \pi_y]$ for all $x$ and $y$.
Lemma 2. $[x,\pi_y,z]$ for all $x,y,z$.
Proof. Suppose we want to try to infer $[x,y,z] = (x\mid y/z) \mid \overline{z/y \mid x}$ from the axiom. In general, things we infer from the axiom have the form $x/v \mid \overline{u/x}$, which matches $[x,y,z]$ if we have $u = z\mid y$ and $v = y\mid z$. Setting $w = y \mid z$ in the axiom to simplify things, we get
$$
    ((z\mid y) \mid (y \mid z)/(y \mid z)) \mid \pi_y / ((x\mid y/z) \mid \overline{z/y \mid x})
$$
which simplifies to
$$
    [z,y] \mid \pi_y / [x,y,z]
$$
so if we prove $[z,y]$, we can infer $[x,y,z]$. In particular, by Lemma 1, we can infer $[x, \pi_y, z]$ for all $x,y,z$.
Theorem. $\pi_y$, $[x,y]$, and $[x,y,z]$ for all $x,y,z$.
Proof. Setting $u=v=w$ in the axiom gives us $\pi_u \mid \pi_y / [x,u]$ which we now "reflect" by Lemma 2 into $[x,u] / \pi_y \mid \pi_u$ and "swap" by Lemma 1 into $\pi_u \mid [x,u] / \pi_y$. So if we prove $\pi_u$ for any $u$, we can use this to infer $\pi_y$ for all $y$.
But we've already proven $[\pi_y, \pi_y]$ (as a special case of Lemma 1) which can be rewritten as
$$
   [\pi_y, \pi_y] = \pi_y / \pi_y \mid \overline{\pi_y / \pi_y} = \pi_{\pi_y \mid \pi_y}.
$$
From $\pi_{\pi_y \mid \pi_y} \mid [x,\pi_y \mid \pi_y] / \pi_y$ and $\pi_{\pi_y \mid \pi_y}$, we infer $\pi_y$ (for all $y$).
Setting $u=v=w=y$ in our earlier formula $(u \mid u/w) \mid \pi_y / [x,u]$ gives us $\pi_y \mid \pi_y / [x,y]$, and now that we know $\pi_y$, we infer $[x,y]$ (for all $x,y$).
Now, from the formula $[z,y] \mid \pi_y / [x,y,z]$, we infer $[x,y,z]$ (for all $x,y,z$).