Realizability for Constructive Zermelo-Fraenkel Set Theory

the background universe, there is a set D such that∀〈u, x〉 ∈ a ∃y ∈ V(A) [〈p(gu)u, y〉 ∈ D ∧ gu ⊩ ϕ(x, y)], and (15)∀z ∈ D ∃〈u, x〉 ∈ a ∃y ∈ V(A) [z = 〈p(gu)u, y〉 ∧ gu ⊩ ϕ(x, y)]. (16)In particular, D ⊆ |A| × V(A), so that by Lemma 3.5, D ∈ V(A). We need to constructe, e ′ ∈ V(A) from g such thate ⊩ ∀x ∈ a ∃y ∈ D ϕ(x, y), (17)e ′ ⊩ ∀y ∈ D ∃x ∈ a ϕ(x, y). (18)For (17), let 〈u, x〉 ∈ a. Then there exists a y such that 〈p(gu)u, y〉 ∈ D and gu ⊩ ϕ(x, y),and hence p(p(gu)u)(gu) ⊩ ∃y ∈ D ϕ(x, y); so that, with e = λu.p(p(gu)u)(gu), (17)obtains.To show (18), let 〈v, y〉 ∈ D. Then, by (16), v = p(gu)u for some u ∈ V(A) and thereexists an x such that 〈u, x〉 ∈ a ∧ gu ⊩ ϕ(x, y). Hence p(v) 1 (v) 0 ⊩ ∃x ∈ a ϕ(x, y), so thatwith e ′ = λv.p(v) 1 (v) 0 , (18) obtains.(Subset Collection): Let a, b ∈ V(A) and ϕ(x, u, y) be a formula with parameters inV(A). We would like to find a realizer r such thatr ⊩ ∃q∀u[∀x ∈ a ∃y ∈ b ϕ(x, y, u) → ∃v ∈ q ϕ ′ (a, v, u)], (19)where ϕ ′ (a, v, u) abbreviates the formulaSetb ∗∀x ∈ a ∃y ∈ v ϕ(x, y, u) ∧ ∀y ∈ v ∃x ∈ a ϕ(x, y, u).= {〈pef, d〉 : e, f ∈ |A| ∧ ef ↓ ∧ 〈(ef) 0 , d〉 ∈ b}.Note that b ∗ is a set. Further, let ψ(e, f, c, u, z) be the formulau ∈ V(A) ∧ e, f ∈ |A| ∧ ef ↓ ∧ ∃d [〈pef, d〉 = z ∧ 〈(ef) 0 , d〉 ∈ b ∧ (ef) 1 ⊩ ϕ(c, d, u)].By invoking Subset Collection there exists a set D such thatNow set∀u∀e [∀ 〈f, c〉 ∈ a ∃z ∈ b ∗ ψ(e, f, c, u, z) →∃w ∈ D (∀ 〈f, c〉 ∈ a ∃z ∈ w ψ(e, f, c, u, z) ∧ ∀z ∈ w ∃ 〈f, c〉 ∈ a ψ(e, f, c, u, z))].Then D ∗ ⊆ V(A), and thusis an element of V(A).Let e ∈ |A| and u ∈ V(A) satisfyD ∗ := {w ∩ b ∗ : w ∈ D}.E := {〈0, v〉 : v ∈ D ∗ }e ⊩ ∀x ∈ a ∃y ∈ b ϕ(x, y, u). (20)16