Lecture 4: Cut-elimination for MLL
Teacher: Michele Pagani
Reminder:
Consistency ⟺
If you have a proof
On top of that, you can prove any
But not with
How to show that
MLL Proof nets
Semi-distributivity of tensor over parr
NB: we don’t have the converse
cf. picture
Now, you have to check if this proof structure is a proof net:
-
4 possible switching graph (erase first left, second left / first left, second right / etc…)
-
in each case: no cycle + connected graph ⟹ we do have a proof net
You have several possible permutation (here: only the identity) to determine axioms. Cut-elimination = composition of permutations
NB: with additives or exponentials: much more challenging, as you can introduce erase some fomulas (ex: as with
- Link:
-
a node, and all its premises and conclusions
cf. picture
NB: the correctness criterion for units is in NPTIME, that’s why we don’t consider units
Cut-elimination for MLL
Th: If there exists a MLL proof nets
, then there exists a cut-free proof net
cf. picture
Cut-elimination rewriting: reduction of ax-redexes and
For ax-redexes: note that we’re only considering proof nets, so there’s no cycle (therefore
cf. picture
Proof net rewriting preserve interfaces (input/output formulas).
Example: cf. picture
NB:
-
this rewriting is confluent, so the cut-free proof obtained is independent of the rewriting path used
-
if you associate a permutation to each proof net, cut-elimination corresponds to composition of permutations (correctness corresponds to properties of the permutations ⟶ geometry of interaction (
-terms are over spaces) (Olivier Laurent)) cf. picture -
normal forms: cycle ax-cut, or cut-free proof nets
Reminder on rewriting systems
- Weak normalization:
-
st is a normal form:
- Strong normalization:
-
st and
NB: this amounts to show that the order induced by
Ex:
is weakly normalizing but not strongly normalizing.
Lemma: the cut-elimination rewriting over MLL proof structure is strongly normalizing
NB:
- the problem is open for MELL
- weakly normalizing for MELL: way harder
Idea: By setting
which will yield the result, as
Warning: we have to prove that
Lemma: Given a reduction step
, if is a proof net, then is a proof net as well.
Sketch: show that if there is a switching cycle with the contractum, there is one as well in the proof net with the redex. And then use the following lemma:
Lemma: if
is an acyclic undirected graph, the the number of connected components if equals
Confluence (Church-Rosser property): the cut-elimination rewriting is confluent.
Because if you have two cuts:
- for
/ -redexes: reducing one cut doesn’t change/touch the other one - pay attention to ax-redexes that overlap: cf. picture (the final result yields the same graph)
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