On Intuitionistic Linear Logic - Microsoft Research

§1. Sequent Calculus 15
Let Π ′ be the proof obtained by applying the inductive hypothesis to the proof
π1
!Γ − A Promotion
!Γ − !A
!Γ n , ∆ − B
π2
!A n , ∆ − B Cutn.
We have by assumption that c(Π1), c(π2) ≤ |!A| and hence c(Π ′ ) ≤ |!A|. Finally take the
proof, Π,
Π ′
!Γ n , ∆ − B Weakening ∗ .
!Γ n+1 , ∆ − B
Hence by definition c(Π) ≤ |!A| and we are done.
(j) (Promotion,Contraction)-cut.
π1
!Γ − A Promotion
!Γ − !A
!Γ n+1 , ∆ − B
π2
!A, !A, !A n , ∆ − B Contraction
!A n+1 , ∆ − B Cutn+1
Let Π ′ be the proof obtained by applying the inductive hypothesis to the proof
π1
!Γ − A Promotion
!Γ − !A
!Γ n+2 , ∆ − B
π2
!A n+2 , ∆ − B Cutn+2.
We have by assumption that c(Π1), c(π2) ≤ |!A| and hence c(Π ′ ) ≤ |!A|. Finally, let Π
be the proof
Π ′
!Γ n+2 , ∆ − B Contraction ∗ .
!Γ n+1 , ∆ − B
Hence by definition c(Π) ≤ |!A| and we are done.
2. When in the proof Π2 the cut formula, A, is a minor formula. We shall consider each case of
the last rule applied in Π2. The case where the cut formula, A, is a minor formula in proof
Π1 is symmetric (and omitted).
(a) tR.
Π1
Γ − A
We can form the (cut-free) proof
Γ n , ∆ − t
(tR)
A n , ∆ − t Cutn
(tR),
Γ n , ∆ − t
which has a cut-rank of 0 and so we are done.