E. If, S. |=-e:-e, and S. |=, ? and |= s : S and e/s a; k ? r

S. If and . |=, ? and S |= k :: ? and |= s : S and s v k ? r

. Proof, The proof is a simple, but tedious induction on the height of the evaluation derivation (the derivation of e/s a; k ? r for (1); the derivation of s v k ? r for (2)), and case analysis on a and k, respectively. We give the main cases; the remaining cases are similar

S. By, S. |=-k, and . |=, By hypothesis S |= v : ? , the value v is a closure (x, a 0 , e 0 ), and (3) E 0 ?x. a 0 : ? 1 ? ? 2 and (4) S e 0 : E 0 for some E 0 . Since v is a closure, the elimination rule for apply1 closures matches. From (3) and (4), we get S |= apply2(x, a 0 , e 0 , k) :: ? 1 . The expected result follows from induction hypothesis

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