TheoremBase

Membership in B is checked value by value, and the predicative expressibility requirement on a typed term does not mention its type.

Proof

We check the two conditions of Terms and Typed Terms under a Guard (ii) for the class BB.

Condition Terms and Typed Terms under a Guard (iii): take an assignment of sets to x1,…,xkx_{1},\dots,x_{k} satisfying γ\gamma. Since tt is of type AA under γ\gamma, the same condition for AA shows that the value of tt under this assignment is a set, and by the hypothesis of (i) this value lies in BB.

Condition Terms and Typed Terms under a Guard (iv) does not mention the type: it asks that, for every set variable zz that is none of x1,…,xkx_{1},\dots,x_{k}, is not free in γ\gamma and does not occur in tt, the condition γ∧z=t\gamma\wedge z=t on x1,…,xkx_{1},\dots,x_{k} and zz be expressed by a predicative formula. This is literally the requirement that the same condition imposes for AA, and it holds because tt is of type AA under γ\gamma.

Both conditions hold for BB, so tt is a term of type BB under γ\gamma.

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