Can all math be done with axioms of length 3 or less?

Question

[1] Assume math can be done with unrestricted grammars (See Are axioms in math equivalent to production rules in unrestricted grammars?).

Can the grammar rules be rewritten to contain at most 3 symbols?

Here's an example:

Suppose we have a grammar rule abcd>efgh.

We can rewrite that grammar rule with these grammar rules

ab>i cd>j ij>k l>ef m>gh n>lm k>n

[2] Now each rule has at most 4 symbols (including the > symbol).

[3] For each rule that has 2 symbols on one side, we could introduce a new bidirectional (=) rule. For example:

Rewrite ab>c as these two rules d>c ab=d

[4] Thus we can rewrite rules:

n>i n=ab o>j o=cd p>k p=ij l>q q=ef m>r r=gh n>s s=lm k>n

[5] We can omit the > and = symbols; the symbols can be inferred. It's always in position 2. We can infer = or > based on the number of symbols. If there are 2 symbols it's >. If there are 3 symbols it's =. We can write our rules as:

ni nab oj ocd pk pij lq qef mr rgh ns slm kn

[5] Therefore any unrestricted grammar can be rewritten as axioms with at most 3 symbols.

[6] By [1] and [5], math is possible with axioms of length 3 or less. (I have not proved the "meta-axioms" to be length 3 or less.)

Is this right?

There appears to be a polynomial overhead to do this:

[7] Suppose we have n total symbols in our axioms (total axiom length).

[8] Each axiom has n or less symbols.

[9] Each axiom can be expressed as a single a>b rule and 0 or more a=bc rules.

How many a=bc rules do we need per axiom?

[10] The a=bc rules can be created as a tree (one tree for each side of a>b):

(For simple arithmetic, assume n is a power of 2; n=2^k.)

[11] Each side of the tree will have at most n symbols.

log(n) + log(n/2) + log(n/4) + ... + log(1)
= log(n/2^0) + log(n/2^1) + log(n/2^2) + ... + log(n/^k)
= log(n) - log(2^0) + log(n) - log(2^1) + log(n) - log(2^2) + ... + log(n) - log(2^k)
= log(n^log(n)) - log(2^0) - log(2^1) - log(2^2) - ... - log(2^k)
= n - log(2^0) - log(2^1) - log(2^2) - ... - log(2^k)
<= n

[12] Therefore, to adding a=bc rules for one side of an a>b rule will create at most log(n) + log(n/2) + log(n/4) + ... + log(1) <= n additional rules.

[13] Therefore, in terms of n, the total axiom length, there is polynomial overhead to have max length of 3 per axiom.

Answer 1

If we allow infinitely many symbols and rules this is trivially doable. Pick your favorite formalization of a sufficiently rich system. We can recursively introduce additional symbols and $4$-symbol rules which reduce each string $\sigma$ in that formalization to a single symbol $c_\sigma$. Now for every valid deduction $\sigma\vdash\tau$ introduce a new rule $c_\sigma\leadsto c_\tau$.

If we want only finitely many symbols and rules, things are a bit more complicated but there's still a fairly boring solution. There are length-four-rule systems which are appropriately Turing complete, and so in a sense we can embed first-order logic into it (specifically, there is a computable way to assign a string $s_\varphi$ in the system to a first-order sentence $\varphi$ such that $s_\varphi\leadsto s_\psi$ iff $\varphi\vdash\psi$).