I tried to post the following as a question (with previous edits) on the logic section of Mathoverflow.  They've put it on hold as unclear as to what I'm asking.  Is Mathoverflow even open to such a question, which has stood as open for a long time, getting asked in the first place? 

Title:

What is the Shortest Axiom of Classical Conditional-Negation Propositional Calculus?

Suppose that we only have propositional variables and connectives. Suppose our rules of inference are detachment {Cαβ, α} ⊢ β, and uniform substitution. Suppose that we have a binary connective "C" and a unary connective "N". In 1953 C. A. Meredith found what is currently the shortest single axiom of classical C-N (Conditional-Negation) propositional calculus [p. 302 of A. N. Prior's Formal Logic indicates this and references Prior's paper " 'Single Axioms for the Systems (C, N), (C, O), and (A, N) of the Two-valued Propositional Calculus' JCS, vol. i, No. 3 (July 1953), pp. 155-64. Systems 1.5, 3.13, 6.13, 6.14.):

CCCCCpqCNrNsrtCCtpCsp.

Or using another prefix notation

→→→→→pq→¬r¬srt→→tp→sp.

So given the first notation, the formation rules are:

1. All lower case letters of the Latin alphabet, as well as any lower case letters of the Latin alphabet sub-scripted by Hindu-Arabic numerals are formulas.
2. If x is a formula, then so is Nx.
3. If x and y are formulas, then so is Cxy.
4. Nothing else is a formula in this context.

To determine what is and what is not a formula the following suffices.

1. Assign -1 to all lower case letters, as well as all lower case letters which are sub-scripted.
2. Assign 0 to N.
3. Assign 1 to C.
4. Sum the assigned numbers as you precede from left to right throughout any given string.
5. A string will qualify as a formula if and only if it either starts and ends with -1, or if it starts with a member of {0, 1} and -1 is only reached at the spot corresponding to the last letter of the string, and that string ends with -1.

Now, via the above it turns out that any formula that starts with "C", has its antecedents corresponding to the longest subformulas which have a "0" corresponding to their last letter. So looking at the above formula, we have

C  C  C  C  C  p  q  C  N  r  N  s  r  t  C  C  t  p  C  s  p
|  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
1  2  3  4  5  4  3  4  4  3  3  2  1  0  1  2  1  0  1  0 -1

The longest subformula which ends with the first "t" is thus CCCCpqCNrNsrt. The longest subformula which ends with the second "p" is thus "Ctp", and "s" is the longest subformula corresponding to the third "0" spot. Thus, the main breaks of the formula can get located by the following "|" marks.

C|CCCCpqCNrNsrt|C|Ctp|C|s|p.

Furthermore, the main breaks of CCCCpqCNrNsrt is given by C|CCCpqCNrNsr|t, and so on.
Suppose we have a 2-valued model of C and N given by the following table:

C  0  1  N
0  1  1  1
1  0  1  0

Does there exist any shorter single axiom, which under the rules above allows us to deduce all tautologies, and only those tautologies in the 2-valued model? Does there exist another single axiom of the same length?
If there exists a unique shortest axiom up to re-symbolization of variables and connectives, what is the shortest axiom which allows us to deduce a known axiom system for classical propositional calculus, such as the following axiom set?

1. CpCqp. [p→(q→p)].
2. CCpCqrCCpqCpr. [(p→(q→r))→((p→q)→(p→r))].
3. CCNpNqCqp. [(¬p→¬q)→(q→p)].

If there exists more than one shortest axiom, what is the set of shortest axioms up to re-symbolization of variables and connectives?

The length of an axiom is defined by the number of symbols it has when fully expressed as a well-formed formula.