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, Since A is finite, we can then reason similarly as in the cases of conjunction (?) above. Cases of ? t ??. We have ? 1 : P + 1 , . . . , ? k : P + k , ? : P + , ? : Q + ? ? : P + with ? Pos ?. Since for S ? ? Q + and m ? N we have ? m+1 ?? (S) = {|?

, By induction on the definition of ?. Case of a refinement type {A | ?}. The result follows from monotony of forcing (i.e. that ? is a subobject of A )

, Then we have x = in i ? y for some i = 0, 1 and some y ? ? |T i | such that y ? n T i . By induction hypothesis we get y ? k T i, Assume x ? n T 0 + T 1 and let k ? n

, Assume x ? n U ? T and let k ? n. But given ? ? k and y ? ? |U | such that y ? ? U we have ev ? ?x, y? ? ? T since ? ? n. Case of ?T . Assume x ? n ?T and let k ? n. If k = 1 then we are done since always x ? 1 ?T . Otherwise, k = ? + 1, so that n = m + 1 with ? ? m. Moreover, there is y ? ? T such that x = next ? y and y ? m T . We get y ? ? T by induction hypothesis, so that x ? k ?T . Case of Fix(X ).A . Assume x ? n Fix(X ).A and let k ? n. We have unfold ? x ? n A, Assume x ? n T 0 × T 1 and let k ? n. Then for each i = 0, 1 we have ? i ? x ? n T i , so that ? i ? x ? k T i by induction hypothesis, and it follows that x ? k T 0 × T 1 . Case of U ? T

, For a pure type A and x ? ? A , we have x ? n A for all n > 0. Proof. The proof is by induction on pairs (n, A), using implicitly Lem. C.2 whenever required. Case of 1. Trivial. and the result follows