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Units and second order in linear logic: msg#00004science.mathematics.frogs
Hello Frogs, There are again some thoughts on the problem of the identity of proofs. This time they concern second order linear logic. It is well-known that the units 1, 0, one, bot of linear logic, can be expressed by using second order quantifiers (and multiplicative connectives). We have: 0 = \forall a.a top = \exists a.a 1 = \forall a.-a # a bot = \exists a.-a * a (where the # is the par, and the * is the tensor of linear logic, and -a is the negation of a.) The question that arises now is the following: (*) Should these logical equivalences also be isomorphisms in a category theoretical axiomatization of the logic? I take votes: For 1/bot: YES NO For 0/top: YES NO Seriously, I could not find any axiomatisation for second order MLL, which I could check. But we know that in the category of coherent spaces they are isos. However, there we also have 1=bot and 0=top, which is certainly not the case in the agreed on axiomatization for MLL/MALL (*-autonomous for MLL, initial/terminal object for 0/top) On the other hand, if we look at the sequent calculus (or CoS, or proof nets), it becomes highly questionable whether we should have these isos. Just by doing the exercise of eliminating the cut from the composition of the canonical proofs for the implications, we see that we have to identify the proof id ---------- -a , a # ---------- -a # a 1 --- bot -------------- 1 bot, -a # a * -------------------------------------- 1 * bot, -a # a E -------------------------------------- \exists a.a * -a, -a # a A -------------------------------------- \exists a.a * -a, \forall a. -a # a with the proof id -------- id ------ a, -a -a,a * -------------------------------------- a * -a, -a, a # -------------------------------------- a * -a, -a # a E -------------------------------------- \exists a.a * -a, -a # a A -------------------------------------- \exists a.a * -a, \forall a. -a # a which seems rather weird to me. What do you think? What should the answer to (*) be ? Ciao, Lutz |
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