Heller M, Woodin W.H. (eds.) Infinity. New research frontiers (CUP, 2011)(ISBN 1107003873)(O)(327s)_MAml_

interpretation of z in b + sde 157proofLeft to the reader.Lemma 8.20 Let S be an isomorphism relation from (A, E, R, x) onto (A ′ ,E ′ , R ′ , x ′ ). Let a ∈ A, b ∈ A ′ . Then S(a,b)↔ (∀c)(R(c, a)↔ (∃d)(S(c, d) ∧R ′ (d,b))).S(a,b)↔ (∀d)(R ′ (d,b)↔ (∃c)(S(c, d) ∧ R(c, a))).proof Let S,A,E,R, x, A ′ , E ′ , R ′ , x ′ be as given. Let a ∈ A, b ∈ A ′ . Theforward direction of the equivalence follows immediately from the definition.Now we assume(1) (∀c)(R(c, a) ↔ (∃d)(S(c, d) ∧ R ′ (d,b))).We claim(2) (∀b ∈ A ′ )(∀d)(R ′ (d,b) ↔ (∃c)(S(c, d) ∧ R(c, a)).To see this, suppose R ′ (d,b). Let S(c, d). By (1), R(c, a). Suppose S(c, d) ∧ R(c, a).By (1), let S(c, d ′ ) ∧ R ′ (d ′ ,b). Then E(d,d ′ ),R ′ (d,b). This establishes (2).Let S(a,b ′ ). By (2),(3) (∀d)(R ′ ↔ d,b ′ )((∃c)(S(c, d) ∧ R(c, a)).(4) (∀d)(R ′ (d,b) ↔ (∃c)(S(c, d) ∧ R(c, a)).Hence, E ′ (b, b ′ ),S(a,b).For the second claim, by Lemma 8.19, S −1 is an isomorphism from (A ′ , E ′ , R ′ , x ′ )onto (A, E, R, x). Apply the first claim to S −1 .Lemma 8.21 Let S be an isomorphism relation from (A, E, R, x) onto (A ′ , E ′ ,R ′ , x ′ ). Then S(x,x ′ ).proof Let S,A,E,R, x, A ′ ,E ′ ,R ′ , x ′ be as given. By Lemma 8.18, x isR maximal. Let S(x,y). We claim that y is R ′ maximal. Suppose R ′ (y,z). LetS(w, z). Then R(x,w), violating the R maximality x. By Lemma 8.18, E ′ (y,x ′ ).Hence, S(x,x ′ ).Lemma 8.22 Let S be an isomorphism relation from (A, E, R, x)|y onto (A ′ ,E ′ , R ′ , x ′ )|y ′ .LetS ′ be an isomorphism relation from (A, E, R, x)|z onto (A ′ ,E ′ , R ′ , x ′ )|z ′ . Then for all a ∈ dom(S) ∩ dom(S ′ ), (∀b)(S(a,b) ↔(a,b)).proof Let S,A,E,R, x,y, A ′ , E ′ , R ′ , x ′ , y ′ , S ′ , z, z ′ be as given. Let a be Rminimal such that a ∈ dom(S) ∩ dom(S ′ ), (∃b)(S(a,b) ↔¬S ′ (a,b)). Fix b withS(a,b) ↔¬S ′ (a,b). By Lemma 8.20,(1) S(a,b) ↔ (∀c)(R(c, a) ↔ (∃d)(S(c, d) ∧ R ′ (d,b))).(2) S ′ (a,b) ↔ (∀c)(R(c, a) ↔ (∃d)(S(c, d) ∧ R ′ (d,b))).It suffices to prove that the right sides of (1) and (2) are equivalent. This is clearbecause if R(c, a), then c ∈ dom(S) ∩ dom(S ′ ), and (∀d)(S(c, d) ↔ S ′ (c, d)).Lemma 8.23 Let S be an isomorphism relation from (A, E, R, x) onto (A ′ , E ′ ,R ′ , x ′ )|y ′ and S ′ be an isomorphism relation from (A, E, R, x) onto (A ′ , E ′ , R ′ ,x ′ )|z ′ . Then E(y ′ , z ′ ) and S = S ′ .