On Intuitionistic Linear Logic - Microsoft Research

§1. Sequent Calculus 13
(c) (⊗R, ⊗L)-cut.
π1
Γ − A
Γ, ∆ − A⊗B
π2
∆ − B (⊗R)
Γ n+1 , ∆ n+1 , Θ − C
π3
A, B, (A⊗B) n , Θ − C (⊗L)
(A⊗B) n+1 , Θ − C Cutn+1
Let Π ′ be the proof obtained by applying the inductive hypothesis to the proof
π1
Γ − A
Γ, ∆ − A⊗B
π2
∆ − B (⊗R)
A, B, Γ n , ∆ n , Θ − C
π3
A, B, (A⊗B) n , Θ − C Cutn.
We have by assumption c(Π1), c(π3) ≤ |A⊗B| and hence c(Π ′ ) ≤ |A⊗B|. Now let Π ′′ be
the proof
π1
Γ − A
Π ′
A, B, Γ n , ∆ n , Θ − C Cut1.
B, Γ n+1 , ∆ n , Θ − C
Since we have by assumption that c(π1) ≤ |A⊗B|, then by definition c(Π ′′ ) ≤ max{(|A|+
1), |A⊗B|, |A⊗B|}. Finally, we can form the following proof, Π,
π2
∆ − B
Π ′′
B, Γ n+1 , ∆ n , Θ − C Cut1.
Γ n+1 , ∆ n+1 , Θ − C
Thus by definition c(Π) ≤ max{(|B| + 1), |A⊗B|, |A⊗B|}; hence c(Π) ≤ |A⊗B| and we
are done.
(d) (&R, &L−1)-cut.
π1
Γ − A
Γ − A&B
π2
Γ − B (&R)
Γ n+1 , ∆ − C
π3
∆, (A&B) n , A − C &L−1
∆, (A&B) n+1 − C Cutn+1
Let Π ′ be the proof obtained by applying the inductive hypothesis to the proof
π1
Γ − A
Γ − A&B
π2
Γ − B (&R)
Γ n , ∆ − C
π3
∆, (A&B) n , A − C Cutn.
We have by assumption that c(Π1), c(π3) ≤ |A&B| and hence c(Π ′ |) ≤ |A&B|. We can
form the following proof, Π,
π1
Γ − A
Γ n+1 , ∆ − C
Π ′
Γ n , ∆ − C Cut1.
Thus by definition c(Π ′ ) ≤ max{|A| + 1, |A&B|, |A&B|}; hence c(Π) ≤ |A&B| and we
are done.