Re: (Frogs) Calculus of Structures and substructura
Arno Bastenhof <[email protected]> Fri, 13 Feb 2009 12:33:50 +0100
| Newsgroups | gmane.science.mathematics.frogs |
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| Message-ID | <4e99c7950902130333n291878fdi26873ffc1fa81d49__7892.16659400171$1264448681$gmane$org@mail.gmail.com> |
> Are you saying that the cos-style rules (as I wrote them) are unsound > for your logic? That there is no combination of your numerous > connectives to make them sound That seems to be the case. For instance, take A ^ (B -> C) ----------------- (A -> B) -> C Or, rewritten using the connectives of LG: A * (B\C) |- (A\B)\C Using residuation, we can bring (A\B) over to the other side of the turnstile: (A\B)*(A*(B\C)) |- C But now we need to rearrange the antecedent into (A * (A\B)) * (B\C), for which we need associativity and commutativity (which we don't generally have in LG). Of course, we can also try other formulations, like A * (B\C) |- C/(B/A) or A * (B\C) |- (B/A)\C or A*(B\C) |- C/(A\B), etc. But none of these are provable in LG (or, more generally, Lambek-style calculi) without assuming associativity and/or commutativity of the tensor. Note that the middle one ( A* (B\C) |- (B/A)\C, or, dually, (C/B)*A |- C/(A\B) ) only requires associativity, so maybe that one could still be used for a cos style presentation of the associative (non-commutative) Lambek calculus. As for the more general rule (A -> B) ^ ( C -> D) -------------------------- (B -> C) -> (A -> D) The closest sequent that I can find the provability of which only requires associativity is (A\B) * (C\D) |- A\( (C/B) \ D) But we're not there yet: we haven't yet validated the i-up rule. For that we could assume something like the existence of a unit for the tensor (which is also not something that we would generally want in the case of LG). On Fri, Feb 13, 2009 at 11:34 AM, Kai Brünnler <[email protected]> wrote: > Arno, > >> Regarding the derivation >> of rules like (A /\ (A -> B)) -> B in cos: if you go the other way >> around (trying to derive the the cos rules), just having (A /\ (A -> >> B)) -> B and A -> B -> (B /\ A) won't be enough: you also require the >> conjunction to be at least associative (and probably also commutative, > > Are you saying that the cos-style rules (as I wrote them) are unsound > for your logic? That there is no combination of your numerous > connectives to make them sound? > > -Kai >