One thing I thought reading your file is "you desperately need iff". There is an alarming number of axioms in the file, maybe more than 50%. I'm not completely up on linear logic but you should be able to define a connective that means iff; I think it is (A <-> B) is ( A -o B ) & ( B -o A ). You should introduce this early, give only the axioms you need to characterize lolly and iff, and then use iff to define the other axioms like distributivity of various kinds.
Duality should allow you to cut down on most of the axioms. The postfix _|_ operator is not a negation in the logic, but rather a defined operation on expressions. To encode this in metamath we need a judgment for ~A = B, for example:
$c ~= $.
$a wff ( ph ~= ps ) $. $( this is slightly abusive as we don't really want ~= to be an operator, just a top level judgment. But the alternative requires more sorts $)
$e |- (
ph
~= ps ) $a
|- ( ps ~= ph ) $.
$e |- ( ph ~= ch ) $e |- ( ps ~= th ) $a
|- ( ( ph (+) ps ) ~= ( ch \/ th ) ) $.
$e |- ( ph ~= ps ) $a
|- ( ! ph ~= ? ps ) $.
We still need a negation in the logic to denote the negation of an atom:
$c ~ $.
$a wff ~ ph $.
$a |- ( ph ~= ~ ph ) $.
You might consider this not parsimonious since ~= is provably equvalent to
~
A <-> B once you have the axioms to say that. But you have to either introduce equality or equals-not early on in order to define negation that is used in the axioms.
Linear logic lends itself well to a one-sided sequent calculus, but since we have to model the comma as an operator on wffs it makes sense to use par since the operator already exists. So then it just boils down to writing the axioms of LL where par substitutes for the context disjunction. Since the context disjunction is commutative and associative we need that for par:
$e |- ( ph \/ ps ) $a ( ps \/ ph ) $.
$e |- ( ( ph \/ ps ) \/ ch ) $a ( ph \/ ( ps \/ ch ) ) $.
$e |-
( ph \/ ( ps \/ ch ) )
$a
( ( ph \/ ps ) \/ ch )
$.
These on their own aren't good enough to put any disjunct wherever we want though, as there is no way to apply associativity to a subterm in the context. In classical logic, we would deduce ( ( ps /\ ph ) /\ ch ) from
( ( ph /\ ps ) /\ ch ) by using modus ponens to reduce to (
( ( ph /\ ps ) /\ ch ) ->
( ( ps /\ ph ) /\ ch ) ) and then use an implication lemma to reduce to ( ( ph /\ ps ) -> ( ps /\ ph ) ) which is an application of symmetry. Having equals here would be again very nice, and we know it's coming but maybe we can't wait because we're still setting up initial proof rules.
Let's give in to temptation and add = as a primitive connective to the logic. This means we don't need ~= anymore, and we have the following:
$c <-> $.
$a |- ( ph <-> ps ) $.
$e |- ( ph <-> ps
) $e |- ( ch <-> ps )
$a |- ( ph <-> ps
) $.
$e |- ph $e |- ( ph <-> ps )
$a |- ps
$.
This should eliminate the need for stating most things as inference rules. Let's get back to contexts:
$a |- ( ( ph \/ ps ) <-> ( ps \/ ph ) ) $.
$a |- ( ( ( ph \/ ps ) \/ ch ) <-> ( ph \/ ( ps \/ ch ) ) ) $.
$e |-
( ph <-> ps )
$e |-
( ch <-> th )
$a
( ( ph \/ ps ) <-> ( ch \/ th ) )
$.
Nice, only one of them is an inference rule now. It would be nice to have the last one as a closed form lemma too but that needs the lolly and we aren't there yet (and there are also linearity worries there). In the spirit of parsimony, we can define half of the connectives in terms of negation. (Let's say that \/, &, 1, T, ! are primitive and the others are defined.) Let's put in the other equality axioms too:
$e |-
( ph <-> ps )
$e |-
( ch <-> th )
$a
( ( ph & ps ) <-> ( ch & th ) )
$.
$e |-
( ph <-> ps )
$a
|- ( ~ ph <-> ~ ps )
$.
We should also have an equality axiom for equality itself, but we don't really need to use <-> as a proper wff and once we have the full axiom system it will be provable. The context disjunction now works correctly, so we can get on with the real axioms:
$a |- ( ~ ~ ph <-> ph ) $.
$a |- ( ~ ph \/ ph ) $.
$e |- ( ph \/ ps ) $e |- ( ~ ps \/ ch ) $a |- ( ph \/ ch ) $.
$a |- 1 $.
$a |- ( ph \/ T ) $.
$e |- ( ph \/ ps ) $e |- ( ph \/ ch ) $a |- ( ph \/ ( ps & ch ) ) $.
$e |- ( ~ ph \/ ps ) $e |- ( ~ ps \/ ph ) $a |- ( ph <-> ps ) $.
The only rule we need for negation is DNE because of our choice to define the other connectives, but we add a rule for proving an equality because otherwise we won't be able to characterize it. We define the dual connectives:
$a |- ( ( ph (+) ps ) <-> ~ ( ~ ph & ~ ps ) ) $.
$a |- ( ( ph (x) ps ) <-> ~ ( ~ ph \/ ~ ps ) ) $.
$a |- ( ? ph <-> ~ ! ~ ph ) $.
$a |- ( 0 <-> ~ T ) $.
$a |- ( F <-> ~ 1 ) $.
$a |- ( ( ph -o ps ) <-> ( ~ ph \/ ps ) ) $.
The cost of having these as definitions is that we still need the rules that were associated to them, but now they are rules for negated forms of the original connectives. At least we have the lolly now so these look good.
$e |-
( ph -o ch )
$e |- ( ps -o
th ) $a (
( ph \/ ps ) -o ( ch \/ th ) ) $.
$a ( ( ph & ps ) -o
ph ) $.
$a ( ( ph & ps ) -o ps ) $.
The wikipedia rules for exponentials are a bit weird since they involve both of them at once. Possibly this can be optimized.
$e |- ph $a |- ( ph \/ ? ps ) $.
$e |- ( ph \/ ( ? ps \/ ? ps ) ) $a |- ( ph \/ ? ps ) $.
$e |- ( ? ph \/ ps ) $a |- ( ? ph \/ ! ps ) $.
$e |- ( ph \/ ps ) $a |- ( ph \/ ? ps ) $.
From these, I think all the axioms in your axiomatization are derivable.