Axiom: --------- A |- A Structural Rules: Γ, A, B, Δ |- C Exchange: --------------------------- Γ, B, A, Δ |- C Γ, A, A |- B Contraction: ------------------- Γ, A |- B Γ |- B Weakening: ----------------- Γ, A |- B Logical Rules: Γ, A |- B --> I: ------------------- Γ |- A --> B Γ |- A --> B Γ |- A --> E: ----------------------------------- Γ |- B`A`, `B`, etc. are variables over formulas. Γ, Δ, etc. are variables over (possibly empty) sequences of formulas. Γ `|- A` is a sequent, and is interpreted as claiming that if each of the formulas in Γ is true, then `A` must also be true. This logic allows derivations of theorems like the following:

------- Id A |- A ---------- Weak A, B |- A ------------- --> I A |- B --> A ----------------- --> I |- A --> B --> AShould remind you of simple types. (What was `A --> B --> A` the type of again?) The easy way to grasp the Curry-Howard correspondence is to *label* the proofs. Since we wish to establish a correspondence between this logic and the lambda calculus, the labels will all be terms from the simply-typed lambda calculus. Here are the labeling rules:

Axiom: ----------- x:A |- x:A Structural Rules: Γ, x:A, y:B, Δ |- R:C Exchange: ------------------------------- Γ, y:B, x:A, Δ |- R:C Γ, x:A, x:A |- R:B Contraction: -------------------------- Γ, x:A |- R:B Γ |- R:B Weakening: --------------------- Γ, x:A |- R:B [x chosen fresh] Logical Rules: Γ, x:A |- R:B --> I: ------------------------- Γ |- \xM:A --> B Γ |- f:(A --> B) Γ |- x:A --> E: ------------------------------------- Γ |- (fx):BIn these labeling rules, if a sequence Γ in a premise contains labeled formulas, those labels remain unchanged in the conclusion. What is means for a variable `x` to be chosen *fresh* is that `x` must be distinct from any other variable in any of the labels used in the proof. Using these labeling rules, we can label the proof just given:

------------ Id x:A |- x:A ---------------- Weak x:A, y:B |- x:A ------------------------- --> I x:A |- (\y.x):(B --> A) ---------------------------- --> I |- (\x y. x):A --> B --> AWe have derived the *K* combinator, and typed it at the same time! Need a proof that involves application, and a proof with cut that will show beta reduction, so "normal" proof. [To do: add pairs and destructors; unit and negation...] Excercise: construct a proof whose labeling is the combinator S, something like this: --------- Ax --------- Ax ------- Ax !a --> !a !b --> !b c --> c ----------------------- L-> -------- L! !a,!a->!b --> !b !c --> c --------- Ax ---------------------------------- L-> !a --> !a !a,!b->!c,!a->!b --> c ------------------------------------------ L-> !a,!a,!a->!b->!c,!a->!b --> c ----------------------------- C! !a,!a->!b->!c,!a->!b --> c ------------------------------ L! !a,!a->!b->!c,! (!a->!b) --> c ---------------------------------- L! !a,! (!a->!b->!c),! (!a->!b) --> c ----------------------------------- R! !a,! (!a->!b->!c),! (!a->!b) --> !c ------------------------------------ R-> ! (!a->!b->!c),! (!a->!b) --> !a->!c ------------------------------------- R-> ! (!a->!b) --> ! (!a->!b->!c)->!a->!c --------------------------------------- R-> --> ! (!a->!b)->! (!a->!b->!c)->!a->!c