On Intuitionistic Linear Logic - Microsoft Research

§1. Sequent Calculus 11
2. The cut rank of a proof π, is defined as
c(π) def
⎧
0 If lst(π)=Identity
⎪⎨ c(π1) If lst(π)=Unary rule
= max{c(π1), c(π2)} If lst(π)=Binary rule
⎪⎩
max{|A| + 1, c(π1), c(π2)} If lst(π)=Cut
with Cut formula A.
3. The depth of a proof π, is defined as
d(π) def
=
⎧
⎨
⎩
0 If lst(π)=Identity
d(π1) + 1 If lst(π)=Unary rule
max{d(π1), d(π2)} + 1 If lst(π)=Binary rule.
To facilitate the proof of cut-elimination 3 we find it convenient to replace the Cut rule with an
indexed cut rule
n
Γ − A A, . . . , A, ∆ − B
Cutn.
Γ, . . . , Γ,
∆ − B
n
It is clear that this is a derived rule 4 as it represents the sequence of Cut rules
Γ − A
n
Γ − A A, A, . . . , A,∆ − B
Cut
Γ, A, . . .,A,∆ − B
·
Γ, Γ, . . .,A, ∆ − B
Cut.
Γ, Γ, . . . , Γ,
∆ − B
n
Of course when n = 1 this rule is just the familiar Cut rule as before. It is clear that if we can
prove a cut elimination theorem for this more powerful cut rule, then we have as a corollary the cut
elimination theorem for the standard unary rule. (In what follows we shall use the abbreviation An to represent the sequence A, . . . , A.)
n
Lemma 1. Let Π1 be a proof of Γ − A and Π2 be a proof of ∆, A n − B and assume that
c(Π1), c(Π2) ≤ |A|. A proof, Π, of Γ n , ∆ − B can be constructed such that c(Π) ≤ |A|.
Proof. We proceed by induction on the sum of the depths of the two proofs, i.e. on d(Π1) +d(Π2),
where Π1 and Π2 are the immediate subtrees of the proof
Π1
Γ − A
Γ n , ∆ − B
Π2
A n , ∆ − B Cutn.
There are numerous cases depending on the structure of Π1 and Π2.
1. When the principal formula in the proofs Π1 and Π2 is the cut formula, A.
3 In particular, without this trick of taking a multi-cut rule it seems difficult to find an inductive count that
decreases if we keep a ‘single’ cut rule. It should be possible but, as far as my research has found, it seems to have
eluded proof theorists so far.
4 As it is a derived rule, it is simple to extend the notion of cut rank to handle an occurrence of the Cutn rule.