Type Inference With Simple Subtypes

ion clause, we use A -{x:σ} to denote the set difference, i.e., the type assignment defined by removing x:σ from A. G(x) = { s⊆t }, {x:s} ⊃ x:t G(MN) = let C1, A1 ⊃ M:σ = G(M) C2, A2 ⊃ N:τ = G(N), with type variables renamed to be disjoint from those in G(M) S = UNIFY({α=β | x:α∈A1 and x:β∈A2} ∪ {σ=τ→t}) where t is a fresh type variable in SC1∪SC2∪{St⊆u}, SA1∪SA2 ⊃ MN:u where u is a fresh type variable G(λx.M) = let C, A ⊃ M:τ = G(M) in if x:σ∈A for some σ then C∪{σ→τ⊆u}, (A -{x:σ}) ⊃ λx.M:u else C∪{s→τ⊆u}, A ⊃ λx.M:u, where s, u are fresh type variables The algorithm could conceivably fail in the application case if the call to UNIFY fails. Howev er, we will see that this does not happen. It is not to hard to prove that if G(M) succeeds, then it produces a provable typing for M. THEOREM 13. If G(M) = C,A ⊃ M:σ, then − C,A ⊃ M:σ. It follows, by Lemma 11, that every instance of G(M) is a provable typing for M. Conversely, every provable typing for M is an instance of G(M). THEOREM 14. Suppose − C,A ⊃ M:σ. Then G(M) succeeds and produces a typing with C,A ⊃ M:σ as an instance. Both theorems are proved below. From Lemma 2, we know that every term has a provable CC typing. Therefore, Theorem 14 implies G(M) always succeeds.