User:RJGray/Math

Ordinals
$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Ord(X) \;\And\; u \in X \implies Seg_E(X,u) = u. \\ &\textbf{Proof. } \;\;Pred_E(u) = u \text{ by definition of } E. \;\;u \subseteq X \text{ since } Trans(X). \\ &\therefore Seg_E(X,u) = X \cap Pred_E(u) = X \cap u = u.\\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Ord(Y) \;\And\; X \subset Y \;\And\; Trans(X) \implies X \in Y. \\ &\textbf{Proof. } \;\;\text{Let } u = least(Y \setminus X). \\ &x \in u \implies x \in X: \;\;\text{Since } u = least(Y \setminus X), \, x \in X. \\ &x \in X \implies x \in u: \;\;\text{Let } x \in X. \;\;\text{ If } x = u, \text{ then } x \notin X. \;\;\text{ Contradiction.} \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\!\text{If } u \in x, \text{ then } Trans(X) \implies u \in X. \;\;\text{ Contradiction.} \;\;\therefore x \in u. \\ &\text{By extensionality, } X = u. \;\;\therefore X \in Y.\\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Ord(X) \;\And\; Ord(Y) \implies \text{Only one of the following holds: } X \in Y,\, X = Y,\, Y \in X. \\ &\textbf{Proof. } \;\;Trans(X) \;\And\; Trans(Y) \implies Trans(X \cap Y). \\ &\text{Assume } X \nsubseteq Y \;\And\; Y \nsubseteq X. \;\;\text{ Then } X \cap Y \neq X \;\And\; X \cap Y \neq Y. \;\text{ Hence, } X \cap Y \subset X \;\And\; X \cap Y \subset Y. \\ &\text{By the previous theorem, } Ord(X) \;\And\; Trans(X \cap Y) \;\And\; X \cap Y \subset X \implies X \cap Y \in X. \\ &\text{Similarly, } X \cap Y \in Y. \text{ Thus, }  X \cap Y \in X \cap Y, \text{ which contradicts regularity}. \\ &\therefore X \subseteq Y \lor Y \subseteq X, \text{ which implies } X \subset Y \lor X = Y \lor Y \subset X. \\ &\text{There are 3 ways for 2 of these 3 to be true: } (X \subset Y \;\And\; X = Y) \lor (X = Y \;\And\; Y \subset X) \lor (X \subset Y \;\And\; Y \subset X). \\ &\text{Each of these implies } X \subset X, \text{ which is a contradiction. Therefore, only one of the three holds.} \\ &\text{By the previous theorem, this implies that only one of the following holds: } X \in Y,\, X = Y,\, Y \in X. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;E\, Connex\, On. \\ &\textbf{Proof. } \;\;\text{Implied by the previous theorem.} \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Ord(A) \;\And\; X \in A \implies Ord(X). \\ &\textbf{Proof. } \;\;Ord(A) \implies X \subseteq A. \quad \therefore E\, Well\, A \implies E\, Well\, X. \\ &\text{Let } Z \in Y \in X \in A. \quad Trans(A) \implies Z, Y, X \in A. \\ &E\, Well\, A \;\And\; Z \in Y \in X \implies Z \in X. \quad \therefore Trans(X). \\ &\therefore E\, Well\, X \;\And\; Trans(X) \implies Ord(X). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Ord(A) \implies A \subseteq On. \\ &\textbf{Proof. } \;\;\text{Let } X \in A. \;\text{ Since } Ord(A), \text{ the previous theorem implies } Ord(X). \\ &\text{Since } X \text{ is a set, } X \in On. \quad \therefore A \subseteq On. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Trans(On). \\ &\textbf{Proof. } \;\;\text{Since } A \in On \text{ implies } Ord(A), \text{ the previous theorem implies } A \subseteq On. \quad \therefore Trans(On). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Ord(On). \\ &\textbf{Proof. } \;\;\text{Follows from earlier theorems that imply } E\, Connex\, On \;\And\; Trans(On). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;\forall X(X \subseteq On \implies E\, Well\, X). \\ &\textbf{Proof. } \;\;Ord(On) \implies E\, Well\, On \implies E\, Well\, X \text{ for all } X \subseteq On. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;\forall x(x \in A \implies Trans(x)) \implies Trans(\cup A). \\ &\textbf{Proof. } \;\;\text{Let } u \in v \in \cup A. \;\text{ There exists a } w \text{ such that } v \in w \in A. \\ &Trans(w) \;\And\; u \in v \in w \implies u \in w. \\ &\text{Since } u \in w \in A \implies u \in \cup A, \text{ we have } Trans(\cup A). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;A \subseteq T \;\And\; Trans(T) \implies \cup A \subseteq T. \\ &\textbf{Proof. } \;\;\text{Let } u \in \cup A. \;\text{ Then there exists a } v \text{ such that } u \in v \in A. \\ &A \subseteq T \implies u \in v \in T. \;\;Trans(T) \implies u \in T. \quad \therefore \cup A \subseteq T. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;A \subseteq On \implies Ord(\cup A). \\ &\textbf{Proof. } \;\;\text{Use last two theorems. First implies} \;Trans(\cup A) \text{ since } \forall x[x \in A \subseteq On \implies Trans(x)]. \\ &\text{Second implies} \;E \, Well \cup\! A \text{ since } A \subseteq On \;\And\; Trans(On) \implies \cup A \subseteq On. \\ &\therefore Ord(\cup A). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;\alpha + 1 = \alpha \cup \{\alpha\} \text{ is the ordinal that follows } \alpha. \\ &\textbf{Proof. } \;\;Trans(\alpha + 1): \beta \in \alpha \cup \{\alpha\} \implies \beta \in \alpha \lor \beta = \alpha. \\ &\quad\text{Now }\beta \in \alpha \implies \beta \subseteq \alpha \text{ and } \beta = \alpha \implies \beta \subseteq \{\alpha\}. \quad\therefore \beta \subseteq \alpha \cup \{\alpha\}. \\ &E\,Well\,\alpha + 1 \text{ since we have just added one set to the end of the well-ordered set }\alpha. \quad \therefore Ord(\alpha +1). \\ &\text{Next we prove: } \alpha < \beta \implies \alpha + 1 \le \beta. \\ &\alpha \in \beta \implies \alpha \subseteq \beta. \;\text{ Also, } \alpha \in \beta \implies \{\alpha\} \subseteq \beta \quad \therefore\alpha \cup \{\alpha\} \subseteq \beta.\\ &\text{This implies } \alpha \cup \{\alpha\} \in \beta \lor \alpha \cup \{\alpha\} = \beta. \quad \therefore \alpha \cup \{\alpha\} \le \beta. \\ &\text{Gödel proves } \beta \in \alpha + 1 \implies \beta \le \alpha: \beta \in \alpha \cup \{\alpha\} \implies \beta \in \alpha \lor \beta = \alpha \implies \beta \le \alpha. \\\\ \end{alignat}$$

Functions on Ordinals
$$\begin{alignat}{2} &\textbf{Theorem. } \;\;\text{If } \beta > \alpha \implies F(\beta) > F(\alpha), \text{ then } \forall \alpha(F(\alpha) \ge \alpha). \\ &\textbf{Proof. } \;\;\text{By contradiction. Assume } \exists \alpha(F(\alpha) < \alpha) \;\text{ and let } \alpha = least\,\{\alpha: F(\alpha) < \alpha\}. \;\text{ Then } F(\alpha) = \beta < \alpha. \\ &\text{However, } F(\beta) \ge \beta \;\And\; F(\alpha) > F(\beta) \implies F(\alpha) > \beta. \;\text{ Contradiction.}\\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Ord(X) \;\And\; Ord(Y) \;\And\; F\,Isom_{E,E}(X,Y) \implies X=Y \;\And\; F=I \upharpoonright X. \\ &\textbf{Proof. } \;\;\text{Let } \alpha = least\,\{\alpha \in X: F(\alpha) \ne \alpha\} \;\text{ and let } F(\alpha) = \beta. \\ &\beta < \alpha: \text{Then } F(\alpha) = \beta = F(\beta), \;\text{ contradicting } F \text{ being one-to-one}.\\ &\beta > \alpha: \text{ If } \gamma < \alpha, \text{ then } F(\gamma) = \gamma < \alpha. \;\text{ If } \gamma \ge \alpha, \text{ then } F(\gamma) \ge F(\alpha) = \beta > \alpha, \text{ contradicting } F \text{ being onto.} \\ &\text{Thus, } \forall \alpha \in X(F(\alpha) = \alpha). \;\text{ Since } F \text{ is an isomorphism, } Y = Rng(F) = X \;\And\; F=I \upharpoonright X. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;\forall G\, \exists!F\,[F\,Fn\,On\;\And\; \forall \alpha(F(\alpha) = G(F \upharpoonright \alpha)]. \\ &\textbf{Proof. } \;\;\text{Strategy: Define class } K \text{ consisting of all functions } f \text{ that approximate } F \text{ on } \beta. \\ &\text{Let: } f \in K \iff \exists \beta[f\,Fn\,\beta \;\And\; \forall \alpha(\alpha \in \beta \implies f(\alpha) = G(f \upharpoonright \alpha))] \\ &f_1\,Fn\,\beta_1 \in K \;\And\; f_2\,Fn\,\beta_2 \in K \;\And\; \beta_1 < \beta_2 \implies f_1 \subseteq f_2: \\ &\quad\text{Assume conclusion is false. Let } \alpha = least \{\gamma: f_1(\gamma) \ne f_2(\gamma)\}. \\ &\quad\text{However, } f_1(\alpha) = G(f_1 \upharpoonright \alpha)) = G(f_2 \upharpoonright \alpha)) = f_2(\alpha), \text{contradicting our assumption.} \\ &\quad\therefore f_1 \subseteq f_2. \\ &\forall \alpha(\exists!f(f \in K \;\And\; f\,Fn\,\alpha \;\And\; \forall \beta < \alpha(F(\beta) = G(F \upharpoonright \beta)]): \\ &\quad\text{Assume false and let } \alpha \text{ be least such that it is false. However, since } \alpha \text{ is least, we can define} \\ &\quad\text{a function } f \text{ by: } h = \cup\{g \in K: \beta < \alpha \land g\,Fn\,\beta\}; \;f = h \cup \{\langle \alpha, G(h \upharpoonright \alpha) \rangle\}, \\ &\quad\text{contradicting our assumption.} \\ &\text{Let } F = \cup K. \;\;F\,Fn\,On \;\And\; \forall\alpha(F(\alpha) = G(F \upharpoonright \alpha)): \\ &\quad\text{The first statement proved above implies that } F \text{ is a function satisfying the recursive definition.} \\ &\quad\text{The second statement implies that } Dom(F) = On. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Pr(A) \;\And\; W\,Well\,A \;\And\; \text{All proper initial segments of } A \text{ are sets} \implies Isom_{E,W}(On,A). \\ &\textbf{Proof. } \;\;\text{Define: } F(\alpha) = least(A \setminus F \upharpoonright \alpha). \\ &\text{If } \forall \beta(\beta < \alpha \implies Seg_W(A, F(\beta)) = F \upharpoonright \beta \text{ is a proper initial segment of } A), \\ &\text{then } Seg_W(A, F(\alpha)) = F \upharpoonright \alpha \text{ is a proper initial segment of } A. \\ &\quad\text{Proof by transfinite induction:} \\ &\quad\alpha = 0: Seg_W(A, F(0)) = \empty = F \upharpoonright \empty, \text{ which is a proper initial segment.} \\ &\quad\alpha = \beta + 1: Seg_W(A, F(\alpha)) = Seg_W(A, F(\beta)) \cup \{ F(\beta) \} = F \upharpoonright \beta \cup \{ F(\beta) \} = F \upharpoonright \alpha, \\ &\quad\text{which is a proper initial segment since } F(\beta) \text{ is ordinal following ordinals in } Seg_W(A, F(\beta)). \\ &\quad\alpha = Lim \{\beta: \beta < \alpha\}: Seg_W(A, F(\alpha)) = \cup \{Seg_W(A, F(\beta)): \beta < \alpha\} = \cup \{F \upharpoonright \beta: \beta < \alpha\} = F \upharpoonright \alpha, \\ &\quad\text{which is a proper initial segment since } F(\alpha) \text{ is ordinal following ordinals in } \cup \{Seg_W(A, F(\beta)): \beta < \alpha\}. \\ &F \text{ preserves ordering : } \,\beta < \alpha \implies F(\beta) \,W\, F(\alpha): F(\beta) \in F \upharpoonright \alpha = Seg_W(A, F(\alpha)) \implies F(\beta) \,W\, F(\alpha). \\ &Dom(F) = On: \text{ Let } \alpha \in On. \;\text{ By replacement } F \upharpoonright \alpha \text{ is a set. Since } A \text{ is a proper class, } A \setminus F \upharpoonright \alpha \\ &\quad\text{is a proper class, so it is non-empty. Thus, } \;F(\alpha) = least(A \setminus F \upharpoonright \alpha) \text{ is defined.} \quad\therefore Dom(F) = Ord. \\ &Rng(F) = A: \text{ Assume } Rng(F) \subset A. \;\;\text{ Let } y = least \{x \in A: x \notin Rng(F)\} \text{ and let } X = \{x: x W y\}. \\ &\quad\text{Since } X \text{ is a proper initial segment, it is a set.} \\ &\quad F \text{ is a bijection from } On\text{ to } X, \text{ so } F^{-1} \text{ is a bijection from } X \text{ to } Ord. \\ &\quad\text{By replacement, } F^{-1}[X] = On \text{ is a set, contradicting } On \text{ being a proper class.} \\ &\quad\therefore Rng(F) = A. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;\text{If } a \text{ is a set and } W\,Well\,a, \text{ then } \exists \alpha(\alpha \in On \land Isom_{E,W}(\alpha,a)). \\ &\textbf{Proof. } \;\;\text{Define: } f(\beta) = least(a \setminus f \upharpoonright \beta). \\ &\text{Same proof of properties as in last theorem with the difference that the construction terminates at } \alpha. \\ &\text{Assume it does not terminate at any ordinal number } \alpha. \;\text{Then } f \text{ is a bijection from } On \text{ to a subset of } a. \\ &\text{So } f^{-1} \text{ is a bijection from a subset of } a \text{ to } On. \\ &\text{By replacement, } f^{-1}[a] = On \text{ is a set, contradicting } On \text{ being a proper class.} \\ &\therefore \exists \alpha(\alpha \in On \land Isom_{E,W}(\alpha,a)). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Definition. } \;\;(\alpha, \beta) \;Le\; (\gamma, \delta) \iff \beta < \delta \lor (\beta = \delta \land \alpha < \gamma). \\ &\textbf{Definition. } \;\;(\alpha, \beta) \;R\; (\gamma, \delta) \iff Max(\alpha, \beta) < Max(\gamma, \delta) \lor (Max(\alpha, \beta) = Max(\gamma, \delta) \land (\alpha, \beta) \;Le\; (\gamma, \delta)) \\ &\textbf{Theorem. } \;\;R \text{ well-orders } On^2. \\ &\textbf{Proof. } \;\;R\;Connex\;On^2: \text{ Follows from } Max(\alpha, \beta) < Max(\gamma, \delta), Max(\alpha, \beta) = Max(\gamma, \delta), \\ &\quad\text{ or } Max(\alpha, \beta) > Max(\gamma, \delta); \text{ and similar inequalities holding between } \beta \text{ and } \delta, \text{ and } \alpha \text{ and } \gamma. \\ &R\;Well\;On^2: \text{ Let } U \subseteq On^2. \\ &\quad\text{Let } min_1 = least\,\{Max(\alpha, \beta): (\alpha, \beta) \in U\} \text{ and let } U_1 = \{(\alpha, \beta) \in U: Max(\alpha, \beta) = min_1\}. \\ &\quad\text{Then } \;\forall(\alpha, \beta) \in U_1 (Max(\alpha, \beta) = min_1) \text{ and } \forall(\alpha, \beta) \in U \setminus U_1[Max(\alpha, \beta) > min_1). \\ &\quad\text{Let } min_2 = least\,\{\beta: (\alpha, \beta) \in U_1\} \text{ and let } U_2 = \{(\alpha, \beta) \in U_1: \beta = min_2\}. \\ &\quad\text{Then } \;\forall(\alpha, \beta) \in U_2 (\beta = min_2) \text{ and } \forall(\alpha, \beta) \in U_1 \setminus U_2(\beta > min_2). \\ &\quad\text{Let } min_3 = least\,\{\alpha: (\alpha, \beta) \in U_2\} \text{ and let } U_3 = \{(\alpha, \beta) \in U_2: \alpha = min_3\}. \\ &\quad\text{Then } \;\forall(\alpha, \beta) \in U_3 (\alpha = min_3) \text{ and } \forall(\alpha, \beta) \in U_2 \setminus U_3(\alpha > min_3). \\ &\quad\text{Now } min = (min_3, min_2) \text{ is the only member of } U_3. \\ &\quad\text{Since } min \in U_3 \subseteq U_2 \subseteq U_1 \subseteq U, \text{ the above properties imply } min < (\alpha, \beta) \text{ for all } (\alpha, \beta) \in U \setminus U3. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Pr(On^2) \;\And\; \text{ All proper initial segments of } (On^2, R) \text{ are sets. Therefore, } Isom_{E,R}(On, On^2). \\ &\textbf{Proof. } \;\;Pr(On) \implies Pr(On^2). \\ &\text{Let } S = Seg_R(On^2, (\alpha, \beta)) \text{ be a proper initial segment and let } \delta = Max(\alpha, \beta). \\ &S \subseteq Seg_R(On^2, (\delta, \delta)+1) \subseteq (\delta+1) \times (\delta+1), \text{ which is a set.} \quad\therefore S \text{ is a set.} \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;\text{If }\, F \,Isom_{E,<}(On, On^2) \text{ as defined above and } P = F^{-1}, \text{ then } P((\alpha, \beta)) \ge max(\alpha, \beta). \\ &\textbf{Proof. } \;\;\text{This follows from: } \;F(max(\alpha, \beta)) \le (max(\alpha, \beta), 0) \le (\alpha, \beta) \\ &\qquad\;\text{since applying } P \text{ gives: } \;max(\alpha, \beta) \le P((max(\alpha, \beta), 0)) \le P((\alpha, \beta)). \\ &\text{All we need to prove is: } F(\alpha) \le (\alpha, 0). \\ &\quad\alpha = \beta + 1: F(\beta + 1) = F(\beta) + 1 \le (\alpha, 0) + 1 \le (\alpha + 1, 0). \\ &\quad\alpha = Lim \{\beta: \beta < \alpha\}: \\ &\quad\quad\text{Lemma: } \text{ If } F \text{ is an isomorphism and } a = Lim \{b: b < a\}, \text{ then } F(a) = Lim \{F(b): b < a\}. \\ &\quad\quad\text{Proof: } \text{ If } F \text{ is an isomorphism and } a = Lim \{b: b < a\}, \text{ then } a \text{ can be defined as } \\ &\quad\quad\quad a \ne least(A) \land \neg \exists b (a = b + 1) \land a = least(A \setminus \{b: b < a\}). \text{ Since } F \text{ is an isomorphism,}\\ &\quad\quad\quad F(a) \ne least(F(A)) \land \neg \exists F(b)(F(a) = F(b) + 1) \land F(a) = least(F(A) \setminus \{F(b): F(b) < F(a)\}). \\ &\quad\quad\quad\therefore F(a) = Lim \{F(b): F(b) < F(a)\} = Lim \{F(b): b < a\}. \\ &\quad\quad\therefore F(\alpha) = Lim \{F(\beta): \beta < \alpha\} \le Lim \{(\beta, 0): \beta < \alpha\} = (\alpha, 0). \text{The last equality holds since } \\ &\quad\quad\quad Lim \{(\beta, 0): \beta < \alpha\} = least(On^2 \setminus \cup\{seg_R(On^2, (\beta, 0)): (\beta, 0) \,R\, (\alpha, 0)\}) = (\alpha, 0). \\ &\text{Gödel's proof is much shorter: } P((\alpha, \beta)) \ge P((max(\alpha, \beta), 0)). \;\;\text{Let } \gamma = max(\alpha, \beta).\\ &\quad\text{Because } P((\gamma, 0)) \text{ is strictly increasing, } P((\gamma, 0)) \ge \gamma. \\ &\quad\therefore P((\alpha, \beta)) \ge P((max(\alpha, \beta), 0)) \ge max(\alpha, \beta). \\\\ \end{alignat}$$

Construction of L
$$\begin{alignat}{2} &\textbf{Definition. } \;\;&&(i, \alpha, \beta) \;S\; (j, \gamma, \delta) \iff (\alpha, \beta) \;R\; (\gamma, \delta) \lor ((\alpha, \beta) = (\gamma, \delta) \land i < j. \\\\ &\textbf{Theorem. } &&Pr(8 \times On^2) \;\And\; \text{ All proper initial segments of } (8 \times On^2, S) \text{ are sets. Therefore, } Isom_{E,S}(On, 8 \times On^2). \\ &\textbf{Proof. } &&Pr(On) \implies Pr(8 \times On^2). \\ & &&\text{Let } A = Seg_S(8 \times On^2, (i, \alpha, \beta)) \text{ be a proper initial segment and let } \delta = Max(\alpha, \beta). \\ & &&A \subseteq Seg_S(8 \times On^2, (8, \delta, \delta)+1) \subseteq 8 \times (\delta+1) \times (\delta+1), \text{ which is a set.} \quad\therefore A \text{ is a set.} \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Definition. } \;\;&&J\,Fn\, (8 \times On^2) \land Rng(J) = On \land i,j < 8 \implies [(i,\alpha,\beta)\;S\;(j,\gamma,\delta) \implies J((i,\alpha,\beta)) < J((j,\gamma,\delta))]. \\ &\textbf{Definition. } &&\text{For } 0 \le i \le 7: \,J_i((\alpha,\beta)) = J((i,\alpha,\beta)). \\\\ &\textbf{Theorem. } &&\text{The } Rng(J_i) \text{ are mutually exclusive and their union is } On. \\ &\textbf{Proof. } &&\text{The ranges are mutually exclusive because } J \text{ is one-to-one.} \\ & &&\text{The union of the ranges is } On \text{ because } J \text{ is onto } On. \\\\ \end{alignat}$$

$$\text{For any } \gamma, \text{ there is a unique } (i, \alpha, \beta) \text{ such that } \gamma = J(i, \alpha, \beta).$$ $$\therefore \text{ There are functions } K_1 \text{ and } K_2 \text{ on } On \text{ such that } K_1(J_i(\alpha, \beta)) = \alpha \;\And\; K_2(J_i(\alpha, \beta)) = \beta.$$

$$\begin{alignat}{2} &\textbf{Definition. } \;\;&&(\alpha,\gamma) \in K_1 \iff \exists i < 8 \,\exists \beta\,(\gamma = J((i, \alpha, \beta))) \land K_1 \subseteq On^2. \\ &\textbf{Definition. } &&(\beta,\gamma) \in K_2 \iff \exists i < 8 \,\exists \alpha\,(\gamma = J((i, \alpha, \beta))) \land K_2 \subseteq On^2. \\\\ &\textbf{Theorem. } &&J_i((\alpha, \beta)) \ge max(\alpha, \beta), \quad J_i((\alpha, \beta)) > max(\alpha, \beta) \text{ for } i > 0, \\ & &&K_1(\alpha) \le \alpha, \qquad\qquad\qquad\, K_1(\alpha) < \alpha \text{ for } \alpha \notin Rng(J_0), \\ & &&K_2(\alpha) \le \alpha, \qquad\qquad\qquad\, K_2(\alpha) < \alpha \text{ for } \alpha \notin Rng(J_0). \\ &\textbf{Proof. } &&J((i, \alpha, \beta)) \ge J((i, (max(\alpha, \beta), 0)).\\ & &&\text{Because } J((i, (max(\alpha, \beta), 0)) \text{ is strictly increasing, } J((i, \gamma, 0)) \ge \gamma. \\ & &&\therefore \text{For } i > 0: J((i, \alpha, \beta)) > J((0, \alpha, \beta)) \ge J((0, (max(\alpha, \beta), 0)) \ge max(\alpha, \beta). \\ & &&\text{Since for } i > 0: \;J((i, \alpha, \beta)) > J((0, \alpha, \beta)) \ge \alpha. \\ & &&\text{We have } i > 0: \;K_1(J((i, \alpha, \beta))) > K_1(J((0, \alpha, \beta))) \ge K_1(\alpha). \\ & &&\text{For all } \alpha: K_1(\alpha) \le K_1(J((i, \alpha, \beta))) = \alpha. \\ & &&\text{For all } \alpha \notin Rng(J_0):  i > 0 \land K_1(\alpha) \le K_1(J((0, \alpha, \beta))) < K_1(J((i, \alpha, \beta))) = \alpha. \\ & &&\text{Similarly for } K_2(\alpha). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Definition. } \;\;&&\text{The following definitions are for subscripts } \,i \,\text{ in } \,J_i, \,\mathfrak{F}_i(X, Y), \,q_i \text{ and } \,Q_i. \\ & &&p = 1 \;\,\quad\text{// Pair} \\ & &&e = 2 \;\,\quad\text{// Epsilon} \\ & &&m = 3 \quad\text{// Minus for set minus (set difference)} \\ & &&v = 4 \;\,\quad\text{// Product by V} \\ & &&d = 5 \;\,\quad\text{// Domain} \\ & &&t = 6 \;\,\quad\text{// Transposition} \\ & &&c = 7 \;\,\quad\text{// Circular permutation} \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\text{Define the element functions: } \;&&q_5 = q_d = dom \;\text{ where } dom((y,x)) = x \land dom \,\mathfrak{Fn}\,V^2. \\ & &&q_6 = q_t = trans \;\text{ where } tr((x,y,z)) = (y,x,z) \land tr \,\mathfrak{Fn}\,V^3. \\ & &&q_7 = q_c = cp \;\text{ where } cp((x,y,z)) = (z,x,y) \land cp \,\mathfrak{Fn}\,V^3. \\ & &&tr_2((x,y) = (y,x) \land tr_2 \,\mathfrak{Fn}\,V^2. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\text{Define the class functions: } \;&&Q_5 = Q_d = Dom \;\text{ where } Dom(X) = dom[X]. \\ & &&Q_6 = Q_t = Tr \;\text{ where } Tr(X) = tr[X]. \\ & &&Q_7 = Q_c = Cp \;\text{ where } Cp(X) = cp[X]. \\ & &&Q_4 = Q_v = Pv \;\text{ where } Pv(X) = dom^{-1}[X] = V \times X. \\ & &&\text{NOTE: } \;dom^{-1} \text{ refers to inverse of the relation } dom. \\ & &&Tr_2(X) = tr_2[X]. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\text{Define the set construction operators: } \;\;&&\mathfrak{F}_1(X, Y) = \mathfrak{F}_p(X, Y) = \{X, Y\}, \\ & &&\mathfrak{F}_2(X, Y) = \mathfrak{F}_e(X, Y) = E \cap X, \\ & &&\mathfrak{F}_3(X, Y) = \mathfrak{F}_m(X, Y) = X \setminus Y, \\ & &&\mathfrak{F}_4(X, Y) = \mathfrak{F}_v(X, Y) = X \cap Pv(Y), \;\;\;\;\quad\text{// Product by V} \\ & &&\mathfrak{F}_5(X, Y) = \mathfrak{F}_d(X, Y) = X \cap Dom(Y), \quad\text{// Domain}\\ & &&\mathfrak{F}_6(X, Y) = \mathfrak{F}_t(X, Y) = X \cap Tr(Y), \;\;\;\;\quad\text{// Transposition} \\ & &&\mathfrak{F}_7(X, Y) = \mathfrak{F}_c(X, Y) = X \cap Cp(Y), \;\;\;\quad\text{// Circular permutation} \\\\ \end{alignat}$$

$$\text{In the following, } 1 \le i \le 7.$$

$$\begin{alignat}{2} &\textbf{Definition. } \;\;&&\alpha \in Rng(J_0) \implies F(\alpha) = Rng(F\upharpoonright\alpha), \\ & &&\alpha \in Rng(J_i) \implies F(\alpha) = \mathfrak{F}_i(F(K_1(\alpha)), F(K_2(\alpha))), \\ & &&F \,Fn\, On. \\\\ &\textbf{Theorem. } &&\text{The function } F \text{ exists.} \\ &\textbf{Proof. } &&\text{By transfinite recursion. Define the function } G \subseteq V^2: \\ & &&(x,y) \in G\iff Dom(x) \in Rng(J_0) \implies y = Rng(x), \\ & && \qquad\qquad\;\; \iff Dom(x) \in Rng(J_i) \implies y = \mathfrak{F}_i(x(K_1(Dom(x)), \,x(K_2(Dom(x))), \\ & && \qquad\qquad\;\; \iff \text{otherwise, } y = \empty. \\ & &&\text{So by transfinite recursion, the function } F = G(F\upharpoonright\alpha) \text{ exists.} \\ & &&\text{We now prove this function is the function in the above definition.} \\ & &&\text{We have } Dom(x) = Dom(F\upharpoonright\alpha) = \alpha. \\ & &&\text{Since } i > 0, \text{ for } \alpha \in Rng(J_i), K_1(Dom(x)) = K_1(\alpha) < \alpha \text{ and } K_2(Dom(x)) = K_2(\alpha) < \alpha. \\ & &&\text{This implies } x(K_1(Dom(x)) = F\upharpoonright\alpha(K_1(\alpha)) = F(K_1(\alpha)) \text{ and similarly for } x(K_2(Dom(x)). \\ & &&\therefore \;\;\alpha \in Rng(J_0) \implies F(\alpha) = Rng(F\upharpoonright\alpha), \\ & &&\quad\;\; \alpha \in Rng(J_i) \implies F(\alpha) = \mathfrak{F}_i(F(K_1(\alpha)), F(K_2(\alpha))). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;&&F(J_i(\alpha, \beta)) = \mathfrak{F}_i(K_1(\alpha), K_2(\beta)). \\ &\textbf{Proof. } &&\text{From last theorem: } \;F(J_i(\alpha, \beta)) = \mathfrak{F}_i(F(K_1(J_i(\alpha, \beta))), F(K_2(J_i(\alpha, \beta)))) \\ & &&\qquad\qquad\qquad\qquad\qquad\qquad\qquad = \mathfrak{F}_i(F(\alpha), F(\beta)). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;&&F(J_0(\alpha, \beta)) = Rng(F\upharpoonright J_0(\alpha, \beta)), \\ & &&F(J_1(\alpha, \beta)) = \{F(\alpha), F(\beta)\}, \\ & &&F(J_2(\alpha, \beta)) = E \cap F(\alpha), \\ & &&F(J_3(\alpha, \beta)) = F(\alpha) \setminus F(\beta), \\ & &&F(J_4(\alpha, \beta)) = F(\alpha) \cap Dom(F(\beta)), \\ & &&F(J_5(\alpha, \beta)) = F(\alpha) \cap Pv(F(\beta)), \\ & &&F(J_6(\alpha, \beta)) = F(\alpha) \cap Cp(F(\beta)), \\ & &&F(J_7(\alpha, \beta)) = F(\alpha) \cap Tr(F(\beta)), \\ &\textbf{Proof. } &&\text{Use last theorem and definition of } \mathfrak{F}_i(X, Y). \\\\ &\textbf{Definition. } \;\;&&L = Rng(F). \\ \end{alignat}$$

Program: input x ∈ F(γ), output x = F(δ)
$$\mathbf{function} \;\text{ordinal}(x, \,\gamma)$$ $$\begin{align} \mathbf{input}: \;&x \text{ is a set.} \\ & \gamma \text{ is an ordinal such that } x \in F(\gamma). \\ \;\;\;\;\mathbf{output}: \;&\text{ordinal } \delta \text{ that satisfies: } x = F(\delta) \land \delta < \gamma. \end{align}$$

$$\mathbf{begin}$$
 * $$\gamma = J(i, \alpha, \beta);$$
 * $$\mathbf{case} \;i \;\mathbf{of}$$

$$\begin{alignat}{2} \qquad\quad\;&0: \;\mathbf{if} \;(x = F(\delta) \land \delta < \gamma) \quad\text{ // } x \in F(\gamma) = Rng(F\upharpoonright \gamma) \text{ implies } \exists \delta\,(\delta < \gamma \land x = F(\delta)). \\ &\qquad\quad\,\mathbf{return} \;\delta; \\ &1: \;\mathbf{if} \;x = F(\alpha) \qquad\qquad\quad\text{ // } \alpha, \beta < \gamma. \text{ Also, } x \in F(\gamma) = \{F(\alpha), F(\beta)\} \text{ implies } x = F(\alpha) \lor x = F(\beta). \\ &\qquad\quad\,\mathbf{return} \;\alpha; \\ &\quad\;\;\mathbf{else} \;\mathbf{if} \;x = F(\beta) \\ &\qquad\quad\,\mathbf{return} \;\beta; \\ &2,3,4,5,6,7: \qquad\qquad\quad\;\; \text{ // For some class } Y\!: x \in F(\gamma) = F(\alpha) \cap Y, \text{ so } x \in F(\alpha) \land \alpha < \gamma. \\ &\qquad\quad\,\mathbf{return} \;\text{ordinal}(x, \,\alpha); \\ \end{alignat}$$
 * $$\mathbf{end}$$

$$\mathbf{end}$$

$$\begin{alignat}{2} &\textbf{Theorem. } &&\text{ Function ordinal}(x, \,\gamma) \text{ terminates and returns } \text{ordinal } \delta \text{ that satisfies: } \delta < \gamma \land x = F(\delta). \\ &\textbf{Proof. } &&\text{All cases except the } 2,3,4,5,6,7 \text{ terminate and return an ordinal less than the argument } \gamma. \\ & &&\text{This case does a recursive call with an argument less than } \gamma. \text{ If the function does not terminate,} \\ & &&\text{it is because this case always occurs. This produces an infinite decreasing sequence of ordinals: } \\ & &&\gamma > \gamma_1 > \cdots > \gamma_n > \cdots. \text{ This contradicts the fact that there is no such decreasing sequence of ordinals.} \\ & &&\therefore \;\text{ The function terminates in one of the other cases and returns a } \delta \text{ that satisfies: } \delta < \gamma \land x = F(\delta). \\\\ \end{alignat}$$

Theorems about L
$$\begin{alignat}{2} &\textbf{Definition. } \;\;&&\mathfrak{L}(A) \iff A \subseteq L \land \forall x(x \in L \implies A \cap x \in L). \\ & &&\text{That is, a class is constructible if and only if its members are constructible} \\ & &&\text{and the intersection with any constructible set is constructible.} \\ &\textbf{Definition. } &&\bar{x}, \dots, \bar{z} \text{ will be used as variables for constructible sets and } \\ & &&\bar{X}, \dots, \bar{Z} \text{ will be used as variables for constructible classes.} \\ &\textbf{Definition. } &&\text{The order of set } x \text{ is the smallest ordinal } \alpha \text{ such that } x = F(\alpha). \\ & &&(x, \alpha) \in Od \iff (x, \alpha) \in F \land \forall \beta(\beta \in \alpha \implies (x, \beta) \notin F) \land Od \subseteq V^2. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;&&x \in F(\alpha) \implies \exists \beta\,(\beta < \alpha \land x = F(\beta)). \\ &\textbf{Proof. } &&\text{See program and theorem of last section.} \\\\ &\textbf{Theorem. } \;\;&&Trans(F[\alpha]). \\ &\textbf{Proof. } &&\text{Let } x \in y \in F[\alpha]. \text{ Then } x \in y = F(\beta) \text{ where } \beta < \alpha. \\ & &&\text{So } x \in F(\beta). \text{ By the theorem above: } \,\exists \delta\,(\delta < \beta < \alpha \land x = F(\delta)). \;\;\therefore x \in F[\alpha]. \\\\ &\textbf{Theorem. } \;\;&&Trans(L). \\ &\textbf{Proof. } &&\text{Let } x \in y \in L. \text{ Then } y = F(\alpha) \text{ for some } \alpha. \\ & &&\text{So } x \in F(\alpha). \text{ By the theorem above: } \,\exists \beta\,(x = F(\beta)). \;\;\therefore x \in L. \\\\ &\textbf{Theorem. } &&x \in y \land x,y \in L \implies Od(x) < Od(y). \text{ In other words, } x \in F[\alpha] \implies Od(x) < \alpha. \\ &\textbf{Proof. } &&\text{Let } Od(y) = \alpha. \text{ Then } x \in F(\alpha), \text{ so } \exists \beta\,(\beta < \alpha \land x = F(\beta)). \\ & &&\therefore Od(x) \le \beta < \alpha = Od(y). \\\\ &\textbf{Theorem. } &&\mathfrak{F_i}(\bar{x},\bar{y}) \in L. \\ &\textbf{Proof. } &&\bar{x} = F(\alpha), \bar{y} = F(\beta) \implies F(J_i(\alpha, \beta)) = \mathfrak{F_i}(\bar{x},\bar{y}). \\\\ &\textbf{Theorem. } &&\bar{x} \cap \bar{y} \in L. \\ &\textbf{Proof. } &&\bar{x} \cap \bar{y} = \bar{x} \setminus (\bar{x} \setminus \bar{y}) = \mathfrak{F_3}(\bar{x},\mathfrak{F_3}(\bar{x},\bar{y})). \\\\ &\textbf{Theorem. } &&x,y \in L \iff (x,y) \in L \text{ and } \;x,y,z \in L \iff (x,y,z) \in L. \\ &\textbf{Proof. } &&x,y \in L \land (x, y) = \{\{x\}, \{x,y\}\} \implies (x, y) = \mathfrak{F_1}(x,\mathfrak{F_1}(x,y)). \\ & &&(x, y) \in L \land (x, y) = \{\{x\}, \{x,y\}\} \implies \{x,y\} \in L \implies x,y \in L. \text{ Or use } Trans(L). \\ & &&\text{By induction, true for } n\text{-tuples.} \\\\ &\textbf{Theorem. } &&(x,y) \in L \iff (y,x) \in L \text{ and } \;(x,y,z) \in L \iff (z,x,y) \in L \iff (x,z,y) \in L. \\ &\textbf{Proof. } &&\text{By last theorem}. \\\\ &\textbf{Theorem. } &&dom(\bar{x}), tr(\bar{x}), cp(\bar{x}), tr_2(\bar{x}) \in L. \\ &\textbf{Proof. } &&\text{By last two theorems}. \\ & &&\text{For example, } \bar{x} = (u, v, w) \implies u,v,w \in L \implies tr(\bar{x}) = (u, w, v) \in L. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;\;&&x \subseteq L \implies \exists \bar{y}(x \subseteq \bar{y}). \\ &\textbf{Proof. } &&\text{Let } \alpha = least\,(On \setminus \{\beta: \exists u(u \in x \land \beta = Od(u)\}). \\ & &&\text{Then } \bar{y} = F(J_0(\alpha, 0)) = Rng(F\upharpoonright\alpha) \text{ satisfies theorem.} \\\\ &\textbf{Theorem. } &&\mathfrak{M}(\bar{X}) \implies \bar{X} \in L. \\ &\textbf{Proof. }  &&\mathfrak{L}(\bar{X}) \implies \bar{X} \subseteq L. \;\;\mathfrak{M}(\bar{X}) \text{ and last theorem imply } \exists \bar{y}(\bar{X} \subseteq \bar{y}). \\ & &&\mathfrak{L}(\bar{X}) \implies \bar{X} = \bar{X} \cap \bar{y} \in L. \\\\ &\textbf{Theorem. } &&\mathfrak{L}(\bar{x}). \\ &\textbf{Proof. } &&\bar{x} \subseteq L. \;\bar{x} \cap \bar{y} \in L. \\\\ &\textbf{Theorem. } &&\bar{x} \cup \bar{y} \in L. \\ &\textbf{Proof. } &&\bar{x} \cup \bar{y} \subseteq L \implies \exists \bar{z}(\bar{x} \cup \bar{y} \subseteq \bar{z}). \\ & &&\bar{x} \cup \bar{y} = (\bar{x}^{c(\bar{z})} \cap \bar{y}^{c(\bar{z})})^{c(\bar{z})} = \bar{z} \setminus ((\bar{z} \setminus \bar{x}) \cap (\bar{z} \setminus \bar{y})). \\ & &&\text{Gödel used: } \;\bar{x} \cup \bar{y} = \bar{z} \setminus ((\bar{z} \setminus \bar{x}) \setminus \bar{y}). \\\\ &\textbf{Theorem. } &&\empty \in L \text{ or } 0 \in L. \\ &\textbf{Proof. } &&F(J_0(0,0)) = Rng(F\upharpoonright 0) = \empty. \;\text{Gödel used: } \;\bar{x} \setminus \bar{x} = \empty. \\\\ &\textbf{Theorem. } &&\mathfrak{L}(L). \\ &\textbf{Proof. } &&L \subseteq L. \;L \cap \bar{x} = \bar{x} \in L. \\ \end{alignat}$$

Theorems about L (section 2)
$$\begin{alignat}{2} &\textbf{Theorem. } \;\;&&\mathfrak{L}(E \cap L). \\ &\textbf{Proof. } &&E \cap L \subseteq L. \\ & &&\text{Let } \bar{x} = F(\alpha). \text{ Then } (E \cap L) \cap \bar{x} = E \cap F(\alpha) = F(J_2(\alpha, 0)) \in L. \\\\

&\textbf{Theorem. } &&\mathfrak{L}(\bar{A} \setminus \bar{B}).\\ &\textbf{Proof. } &&\bar{A} \setminus \bar{B} \subseteq \bar{A} \subseteq L. \\ & &&(\bar{A} \setminus \bar{B}) \cap \bar{x} = (\bar{A} \cap \bar{x}) \setminus (\bar{B} \cap \bar{x}) = \bar{y} \setminus \bar{z} \in L. \\ & &&\text{The last expression is the difference of two constructible sets, which is constructible by } \mathfrak{F_3}. \\\\ &\textbf{Theorem. } &&\mathfrak{L}(\bar{A} \cap \bar{B}).\\ &\textbf{Proof. } &&\bar{A} \cap \bar{B} \subseteq \bar{A} \subseteq L. \\ & &&(\bar{A} \cap \bar{B}) \cap \bar{x} = \bar{A} \cap (\bar{B} \cap \bar{x}) = \bar{A} \cap \bar{y} \in L. \\\\ &\textbf{Theorem. } &&\mathfrak{L}(\bar{A} \cup \bar{B}).\\ &\textbf{Proof. } &&\bar{A} \cup \bar{B} \subseteq L. \\ & &&(\bar{A} \cup \bar{B}) \cap \bar{x} = (\bar{B} \cap \bar{x}) \cup (\bar{A} \cap \bar{x}) = \bar{y} \cup \bar{z} \in L. \\\\

&\textbf{Theorem. } &&Dom(\bar{x}) \in L; \;Tr(\bar{x}) \in L; \;Cp(\bar{x}) \in L. \\ &\textbf{Proof. } &&\text{Let } i = d, t, \text{ or } c. \;Q_i(\bar{x}) \text{ is a set by replacement}. \;\,\therefore \exists \bar{y}(Q_i(\bar{x}) \subseteq \bar{y}). \\ & &&\text{Let } F(\alpha) = \bar{y}; F(\beta) = \bar{x}. \text{ Then } F(J_i(\alpha, \beta)) = F(\alpha) \cap Q_i(F(\beta)) = \bar{y} \cap Q_i(\bar{x}) = Q_i(\bar{x}). \\ & &&\therefore Q_i(\bar{x}) \in L. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;Tr_2(\bar{x}) \in L. \\ &\textbf{Proof. } \;\;Tr_2(\bar{x}) = Dom(Tr(L \cap Pv(\bar{x}))): \\ &y \in Dom(Tr(L \cap Pv(\bar{x}))) \iff \exists u,v,y_2,y_3,w(y = (v, u) \land y_2 = (w, v, u) \land y_3 = (w, u, v) \land w \in L \land (u,v) \in \bar{x}) \\ &\qquad\qquad\qquad\qquad\qquad\quad \implies y \in Tr_2(\bar{x}). \\ &y \in Tr_2(\bar{x}) \iff \exists u,v(y = (v,u) \land (u,v) \in \bar{x})) \\ &\qquad\qquad\quad \implies \exists u,v,y_2,y_3,w(y = (v,u) \land w \in L \land y_3 = (w, u, v) \land y_2 = (w, v, u) \land (u,v) \in \bar{x}) \\ &\qquad\qquad\quad \iff y \in Dom(Tr(L \cap Pv(\bar{x})). \\\\

&\textbf{Theorem. } &&Rng(\bar{x}) \in L. \\ &\textbf{Proof. } \;\;Rng(\bar{x}) = Dom(Tr2(\bar{x})): \\ \\\\ \end{alignat}$$

Theorems about L (section 3)
$$\begin{alignat}{2} &\textbf{Theorem. } \;\;&&\mathfrak{L}(Dom(\bar{A})); \;\mathfrak{L}(Tr(\bar{A})); \;\mathfrak{L}(Cp(\bar{A})).\\ &\textbf{Proof. } &&\text{Let } i = d, t, \text{ or } c. \;\,Q_i(\bar{A}) \subseteq L: v \in Q_i(\bar{A}) \implies \exists u\,(v = q_i(u) \land u \in \bar{A} \subseteq L). \\ & &&\text{From a previous theorem: } u \in L \implies q_i(u) = v \in L. \;\,\therefore Q_i(\bar{A}) \subseteq L. \\ & &&Q_i(\bar{A}) \cap \bar{x} \in L: \;\,\text{Strategy: Find a } \bar{z} \text{ such that } Q_i(\bar{z}) \cap \bar{x} = Q_i(\bar{A}) \cap \bar{x}. \\ & &&\text{Then } Q_i(\bar{z}) \cap \bar{x} \in L \text{ since } F(J(i,\alpha,\beta) = \mathfrak{F}_i(F(\alpha), F(\beta)) = \bar{x} \cap Q_i(\bar{z}). \\ & &&\text{For } v \in Q_i(\bar{A}) \cap \bar{x}, \text{ let } \alpha_v = least\,\{\alpha: u \in \bar{A} \land v = q_i(u) \land u = F(\alpha)\}. \\ & &&\text{Let } y = \{F(\alpha_v): v \in \bar{x}\}, \text{ which is a set by replacement.} \\ & &&\text{Let } \exists \bar{b}(y \subseteq \bar{b}). \;\text{Let } \bar{z} = \bar{A} \cap \bar{b}. \;\text{Then } y \subseteq \bar{z} \subseteq \bar{A}.  \\ & &&\bar{z} \subseteq \bar{A} \implies Q_i(\bar{z}) \cap \bar{x} \subseteq Q_i(\bar{A}) \cap \bar{x}. \\ & &&v \in Q_i(\bar{A}) \cap \bar{x} \implies \exists u(u \in \bar{A} \land v = q_i(u)) \\ & &&\qquad\qquad\qquad\;\, \implies u = F(\alpha_v) \implies u \in y \implies u \in \bar{z} \implies v \in  Q(\bar{z}). \\ & &&\text{Hence, } v \in Q_i(\bar{z}) \cap \bar{x}. \,\text{ So } Q_i(\bar{A}) \cap \bar{x} \subseteq Q(\bar{z}) \cap \bar{x}. \\ & &&\therefore Q_i(\bar{A}) \cap \bar{x} = Q_i(\bar{z}) \cap \bar{x} \in L. \\\\

&\textbf{Theorem. } &&L \cap (V \times \bar{A}) = L \times \bar{A}. \\ &\textbf{Proof. } &&x \in L \cap (V \times \bar{A}) \iff x \in L \land x = (u, v) \land u \in V \land v \in \bar{A}. \\ & &&\qquad\qquad\qquad\quad\;\, \iff x = (u, v) \land u, v \in L \land v \in \bar{A} \\ & &&\qquad\qquad\qquad\quad\;\, \iff x = (u, v) \land u \in L \land v \in \bar{A} \\ & &&\qquad\qquad\qquad\quad\;\, \iff x \in L \times \bar{A}. \\\\

&\textbf{Theorem. } &&\mathfrak{L}(L \cap Pv(\bar{A})) \text{ or } \mathfrak{L}(L \cap (V \times \bar{A})) \text{ or } \mathfrak{L}(L \times \bar{A})). \\ &\textbf{Proof. } &&L \cap Pv(\bar{A}) \subseteq L. \\ & && (L \cap Pv(\bar{A})) \cap \bar{x} \in L: \text{In last proof: replace } Q_i(\bar{A}) \text{ with } L \cap Pv(\bar{A}).\\ & &&(L \cap Pv(\bar{A})) \cap \bar{x} \in L: \;\,\text{Strategy: Find a } \bar{z} \text{ such that } Pv(\bar{z}) \cap \bar{x} = (L \cap Pv(\bar{A})) \cap \bar{x}. \\ & &&\text{Then } Pv(\bar{z}) \cap \bar{x} \in L \text{ since } F(J(v,\alpha,\beta) = \mathfrak{F}_v(F(\alpha), F(\beta)) = \bar{x} \cap Pv(\bar{z}). \\ & &&\text{For } v \in (L \cap Pv(\bar{A})) \cap \bar{x}, \text{ let } \alpha_v = least\,\{\alpha: u \in \bar{A} \land \exists w(v = (w, u) \land u = F(\alpha)\}. \\ & &&\text{Let } y = \{F(\alpha_v): v \in \bar{x}\}, \text{ which is a set by replacement.} \\ & &&\text{Let } \exists \bar{b}(y \subseteq \bar{b}). \;\text{Let } \bar{z} = (L \cap Pv(\bar{A})) \cap \bar{b}. \;\text{Then } y \subseteq \bar{z} \subseteq (L \cap Pv(\bar{A})).  \\ & &&\bar{z} \subseteq (L \cap Pv(\bar{A})) \implies Q_i(\bar{z}) \cap \bar{x} \subseteq (L \cap Pv(\bar{A})) \cap \bar{x}. \\ & &&v \in (L \cap Pv(\bar{A})) \cap \bar{x} \implies \exists u(u \in \bar{A} \land \exists w(v = (w, u))). \\ & &&\qquad\qquad\qquad\qquad\quad \implies u = F(\alpha_v) \implies u \in y \implies u \in \bar{z} \implies v \in  Q(\bar{z}). \\ & &&\text{Hence, } v \in Pv(\bar{z}) \cap \bar{x}. \,\text{ So } (L \cap Pv(\bar{A})) \cap \bar{x} \subseteq Q(\bar{z}) \cap \bar{x}. \\ & &&\therefore (L \cap Pv(\bar{A})) \cap \bar{x} = Pv(\bar{z}) \cap \bar{x} \in L. \\\\

&\textbf{Theorem. } &&\mathfrak{L}(Tr_2(\bar{A})). \\ &\textbf{Proof. } &&Tr_2(\bar{A}) = Dom(Tr(L \cap Pv(\bar{A}))). \\ & &&\mathfrak{L}(L \cap Pv(\bar{A})) \implies \mathfrak{L}(Tr(L \cap Pv(\bar{A})) \\ & &&\qquad\qquad\qquad\; \implies \mathfrak{L}(Dom(Tr(L \cap Pv(\bar{A}))) \\ & &&\qquad\qquad\qquad\; \implies \mathfrak{L}(Tr_2(\bar{A})). \\\\

&\textbf{Theorem. } &&\mathfrak{L}(\bar{A} \times \bar{B}). \\ &\textbf{Proof. } &&(\bar{A} \times L) \cap (L \times \bar{B}) = (\bar{A} \cap L) \times (L \cap \bar{B}) = \bar{A} \times \bar{B}. \\ & &&\text{By a previous theorem: } \mathfrak{L}(L \times \bar{B}). \\ & &&\text{Now } \bar{A} \times L = Tr_2(L \times \bar{A}) \implies \mathfrak{L}(\bar{A} \times L). \\ & &&\therefore \;\mathfrak{L}((\bar{A} \times L) \cap (L \times \bar{B})) \text{ or } \mathfrak{L}(\bar{A} \times \bar{B}). \\\\

&\textbf{Theorem. } &&\mathfrak{L}(Rng(\bar{A})). \\ &\textbf{Proof. } &&Rng((\bar{A}) = Dom(Tr_2((\bar{A}))) \,\text{ so result follows from: } \,\mathfrak{L}(Tr_2(\bar{A})), \,\mathfrak{L}(Dom(\bar{A})). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;&&\mathfrak{L}(\bar{A} \upharpoonright \bar{B}). \\ &\textbf{Proof. } &&\bar{A} \upharpoonright \bar{B} = \bar{A} \cap (V \times \bar{B}) = \bar{A} \cap L \cap (V \times \bar{B}) = \bar{A} \cap (L \times \bar{B}), \\ & &&\text{ so result follows from: } \,\mathfrak{L}(L \times \bar{B}), \,\mathfrak{L}(\bar{A} \cap \bar{B}). \\\\

&\textbf{Theorem. } &&\mathfrak{L}(\bar{A}[\bar{B}]). \\ &\textbf{Proof. } &&\bar{A}[\bar{B}] = Rng(\bar{A} \upharpoonright \bar{B}) \,\text{ so result follows from: } \,\mathfrak{L}(\bar{A} \upharpoonright \bar{B}), \,\mathfrak{L}(Rng(\bar{A})). \\\\

&\textbf{Theorem. } &&\mathfrak{L}(\{\bar{X}, \bar{Y}\}). \\ &\textbf{Proof. } &&\text{By definition of ordered pair for classes: } \{\bar{X}, \bar{Y}\} = \empty, \{\bar{X}\}, \{\bar{Y}\}, \text{ or }\{\bar{X}, \bar{Y}\} \\ & &&\text{where only sets appear in braces. Theorem follows from preceding theorems.} \\\\ \end{alignat}$$

Theorems for proving absoluteness
$$\begin{alignat}{2} &\textbf{Absoluteness } \;\textbf{ theorems. } \;\;\text{These theorems are used to prove absoluteness.} \\ &Abs_1\text{:} \;\bigl[\,\forall u(P(u) \implies Q(u)) \land (P(u) \implies u \in L)\,\bigr] \iff \forall \bar{u}(P(\bar{u}) \implies Q(\bar{u})). \\ &Abs_2\text{:} \;\bigl[\,\forall u(P(u) \iff Q(u)) \land (P(u) \implies u \in L) \land (Q(u) \implies u \in L)\,\bigr] \iff \forall \bar{u}(P(\bar{u}) \iff Q(\bar{u})). \\ &Abs_3\text{:} \;\bigl[\,\forall u\,\forall v\,\forall w(P(u,v,w) \implies Q(u,v,w)) \land (P(u,v,w) \implies u,v,w \in L)\,\bigr] \\ &\qquad\;\iff \forall \bar{u}\,\forall \bar{v}\,\forall \bar{w}(P(\bar{u},\bar{v},\bar{w}) \implies Q(\bar{u},\bar{v},\bar{w})). \\ &\textbf{Proof. } \;\;\Longrightarrow \text{direction:} \\ &\quad\forall u(P(u) \implies Q(u)) \implies \forall u(u \in L \implies (P(u) \implies Q(u))) \iff \forall \bar{u}(P(\bar{u}) \implies Q(\bar{u})). \\ &\quad\forall u(P(u) \iff Q(u)) \implies \forall u(u \in L \implies (P(u) \iff Q(u))) \iff \forall \bar{u}(P(\bar{u}) \iff Q(\bar{u})). \\ &\Longleftarrow \text{direction:} \\ &\text{Case 1: } \;\;u \in L. \\ &\quad\forall \bar{u}(P(\bar{u}) \implies Q(\bar{u})) \iff \forall u(u \in L \implies (P(u) \implies Q(u))), \text{ so } u \in L \text{ implies } P(u) \implies Q(u). \\ &\quad\forall \bar{u}(P(\bar{u}) \iff Q(\bar{u})) \iff \forall u(u \in L \implies (P(u) \iff Q(u))), \text{ so } u \in L \text{ implies } P(u) \iff Q(u). \\ &\text{Case 2: } \;\;u \notin L. \\ &\quad \forall u(u \notin L \implies \neg P(u)), \text{ so } u \notin L \text{ implies } P(u) \implies Q(u). \\ &\quad \forall u(u \notin L \implies \neg P(u) \land \neg Q(u)), \text{ so } u \notin L \text{ implies } P(u) \iff Q(u). \\ &\text{Putting both cases together: } \\ &\quad\forall \bar{u}(P(\bar{u}) \implies Q(\bar{u})) \implies \forall u(P(u) \implies Q(u)). \\ &\quad\forall \bar{u}(P(\bar{u}) \iff Q(\bar{u})) \implies \forall u(P(u) \iff Q(u)). \\\\ \end{alignat}$$

Absoluteness
$$\begin{alignat}{2} &\textbf{Theorem. } \;\;&&\bar{X} \subseteq \bar{Y} \iff \bar{X} \subseteq_l \bar{Y}. \\ &\textbf{Proof. } &&\!\bar{X} \subseteq_l \bar{Y} \iff \forall \bar{u}(\bar{u} \in \bar{X} \implies \bar{u} \in \bar{Y}). \\ & &&\bar{X} \subseteq \bar{Y} \iff \forall u(u \in \bar{X} \implies u \in \bar{Y}) \\ & &&\qquad\quad\, \iff \forall \bar{u}(\bar{u} \in \bar{X} \implies \bar{u} \in \bar{Y}) \text{ by } Abs_1 \text{ since } u \in \bar{X} \implies u \in L. \\ & &&\qquad\quad\; \iff \bar{X} \subseteq_l \bar{Y}. \\\\

&\textbf{Theorem.} &&\bar{X} = \bar{Y} \iff \forall \bar{u}(\bar{u} \in \bar{X} \iff \bar{u} \in \bar{Y}). \\ & &&\text{Constructive extensionality:} \\ & &&\quad\text{Two constructive classes are equal iff they have the same constructible elements.} \\ &\textbf{Proof. } &&\bar{X} = \bar{Y} \iff \forall u(u \in \bar{X} \iff u \in \bar{Y}) \\ & &&\qquad\quad\; \iff \forall \bar{u}(\bar{u} \in \bar{X} \implies \bar{u} \in \bar{Y}) \text{ by } Abs_2 \text{ since } u \in \bar{X} \text{ and } u \in \bar{Y} \implies u \in L. \\\\

&\textbf{Theorem. } &&\{\bar{X}, \bar{Y}\}_l = \{\bar{X}, \bar{Y}\}.\\ &\textbf{Proof. } &&\forall \bar{u}(\bar{u} \in \{\bar{X}, \bar{Y}\}_l \iff \bar{u} = \bar{X} \lor \bar{u} = \bar{Y}). \\ & &&\text{A previous theorem proved that } \{\bar{X}, \bar{Y}\} \in L. \\ & &&\text{Using } Abs_2 \text{ since } u \in \{\bar{X}, \bar{Y}\} \implies u \in L \text{ and } u = \bar{X} \lor u = \bar{Y} \implies u \in L: \\ & &&\forall u(u \in \{\bar{X}, \bar{Y}\} \iff u = \bar{X} \lor u = \bar{Y}) \iff \forall \bar{u}(\bar{u} \in \{\bar{X}, \bar{Y}\} \iff \bar{u} = \bar{X} \lor \bar{u} = \bar{Y}), \\ & &&\therefore \;\{\bar{X}, \bar{Y}\}_l = \{\bar{X}, \bar{Y}\} \text{ by constructive extensionality.} \\\\

&\textbf{Theorem. } &&\text{If } A, B \text{ are absolute, then } A_l(B_l(\bar{X})) = A(B(\bar{X})).\\ &\textbf{Proof. } &&\text{Since } A \text{ is absolute and } B_l(\bar{X}) \text{ is constructive, and } B \text{ is absolute:} \\ & &&\qquad A_l(B_l(\bar{X})) = A(B_l(\bar{X})) = A(B(\bar{X})). \\\\

&\textbf{Theorem. } &&(\bar{X}, \bar{Y})_l = (\bar{X}, \bar{Y}). \;\text{ (An example of last proof.)} \\ &\textbf{Proof. } &&(\bar{X}, \bar{Y})_l = \{\{\bar{X}\}_l, \{\bar{X}, \bar{Y}\}_l\}_l = \{\{\bar{X}\}_l, \{\bar{X}, \bar{Y}\}_l\} = \{\{\bar{X}\}, \{\bar{X}, \bar{Y}\}\} = (\bar{X}, \bar{Y}). \\\\

&\textbf{Theorem. } &&\text{If } _{n}A, \,\ldots, \,_{1}A\text{ are absolute, then } _{n}A_l(\cdots (_{1}A_l(\bar{X})) \,= \,_{n}A(\cdots (_{1}A(\bar{X})).\\ &\textbf{Proof. } &&\text{Basis step:} \;_{1}A_l(\bar{X}) = \,_{1}A(\bar{X}).\\ & &&\text{Inductive step: Assume } \;_{n}A_l(\cdots (_{1}A_l(\bar{X})) \,= \,_{n}A(\cdots (_{1}A(\bar{X})). \\ & &&\text{Then: }\;_{n+1}A_l(_{n}A_l(\cdots (_{1}A_l(\bar{X}))) \,= \,_{n+1}A(_{n}A_l(\cdots (_{1}A_l(\bar{X}))) \,= \,_{n+1}A(_{n}A(\cdots (_{1}A(\bar{X}))). \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Theorem. } \;\;&&\mathfrak{M}_l(\bar{X}) \iff \mathfrak{M}(\bar{X}).\\ &\textbf{Proof. } &&\mathfrak{M}_l(\bar{X}) \iff \bar{X} \in L. \\ & &&\bar{X} \in L \implies \mathfrak{M}(\bar{X}) \text{ by axiom A2. } \;\mathfrak{M}(\bar{X}) \implies \bar{X} \in L \text{ by a previous theorem.} \\ & &&\text{Thus, } \bar{X} \in L \iff \mathfrak{M}(\bar{X}). \;\,\therefore \mathfrak{M}_l(\bar{X}) \iff \mathfrak{M}(\bar{X}). \\\\

&\textbf{Theorem. } &&\mathfrak{Pr}_l(\bar{X}) \iff \mathfrak{Pr}(\bar{X}).\\ &\textbf{Proof. } &&\mathfrak{Pr}_l(\bar{X}) \iff \neg \mathfrak{M}_l(\bar{X}) \iff \neg \mathfrak{M}(\bar{X}) \iff \mathfrak{Pr}(\bar{X}). \\\\

&\textbf{Theorem. } &&V_l = L.\\ &\textbf{Proof. } &&\forall \bar{u}(\bar{u} \in V_l). \\ & &&\forall \bar{u}(\bar{u} \in L). \;\text{ Hence, } \forall \bar{u}(\bar{u} \in V_l \iff \bar{u} \in L). \\ & &&\text{Since } L \text{ is constructive, } \;V_l = L \text{ by constructive extensionality.} \\\\

&\textbf{Theorem. } &&\empty \text{ is absolute}.\\ &\textbf{Proof. } &&\forall u(u \notin \empty) \implies \forall \bar{u}(\bar{u} \notin \empty). \;\text{Also, } \forall \bar{u}(\bar{u} \notin \empty_l). \\ & &&\text{Since } \empty \text{ is constructive, } \;\empty_l = \empty \text{ by constructive extensionality.} \\\\

&\textbf{Theorem. } &&Un_l(\bar{F}) \iff Un(\bar{F}).\\ &\textbf{Proof. } &&Un_l(\bar{F}) \iff \forall \bar{u}\,\forall \bar{v}\,\forall \bar{w}((\bar{v}, \bar{u}) \in \bar{F} \land (\bar{w}, \bar{u}) \in \bar{F} \implies \bar{v} = \bar{w}). \\ & &&Un(\bar{F}) \iff \forall u\,\forall v\,\forall w((v, u) \in \bar{F} \land (w, u) \in \bar{F} \implies v = w). \\ & &&\qquad\quad\, \iff \forall \bar{u}\,\forall \bar{v}\,\forall \bar{w}((\bar{v}, \bar{u}) \in \bar{F} \land (\bar{w}, \bar{u}) \in \bar{F} \implies \bar{v} = \bar{w}) \\ & &&\qquad\qquad\qquad \text{ by } Abs_3 \text{ since } (v, u) \in \bar{F} \land (w, u) \in \bar{F} \implies u, v, w \in L \\ & &&\qquad\quad\, \iff Un_l(\bar{F}). \\\\

&\textbf{Theorem. } &&\cup_l \bar{x} = \cup \bar{x}.\\ &\textbf{Proof. } &&\forall \bar{u}(\bar{u} \in \cup_l \bar{x} \iff \exists \bar{v}(\bar{u} \in \bar{v} \land \bar{v} \in \bar{x})). \\ & &&\text{Strategy: form the set } \{(u, v) \in E \cap (L \times \bar{x})\}. \text{ Then replace L with a set.} \\ & &&\cup \bar{x} \subseteq \bar{w}. \;\;\text{Let } \bar{z} = Dom(E \cap (\bar{w} \times \bar{x})). \\ & && u \in \bar{z} \iff u \in Dom(E \cap (\bar{w} \times \bar{x})) \\ & &&\qquad\;\; \iff \exists v((u, v) \in E \land (u, v) \in \bar{w} \times \bar{x}) \\ & &&\qquad\;\; \iff \exists v(u \in v \land u \in \bar{w} \land v \in \bar{x}) \\ & &&\qquad\;\; \iff \exists v(u \in v \land v \in \bar{x}) \;\;\text{ (Since } \exists v(u \in v \land v \in \bar{x}) \implies u \in \cup \bar{x} \subseteq \bar{w}).\\ & &&\qquad\;\; \iff u \in \cup \bar{x} \;\,\text{ and } \iff \exists \bar{v}(\bar{u} \in \bar{v} \land \bar{v} \in \bar{x}) \\ & &&\qquad\qquad\qquad\qquad\qquad\quad\, \iff \bar{u} \in \cup_l \bar{x}. \\ \end{alignat}$$

L is a model of NBG
$$\begin{alignat}{2} &A1_l: &&\mathfrak{L}(\bar{x}). \\ &\textbf{Proof. } \;\;&&\text{A previous theorem.} \\\\

&A2_l: &&\bar{X} \in \bar{Y} \implies \mathfrak{M}_l(\bar{X}). \\ &\textbf{Proof. } &&\text{Using A2 and absoluteness of } \mathfrak{M}: \;\bar{X} \in \bar{Y} \implies \mathfrak{M}(\bar{X}) \implies \mathfrak{M}_l(\bar{X}).\\\\

&A3_l: &&\forall \bar{u}(\bar{u} \in \bar{X} \iff \bar{u} \in \bar{Y}) \implies \bar{X} = \bar{Y}. \\ &\textbf{Proof. } &&\text{By constructive extensionality.} \\\\

&A4_l: &&\forall \bar{x} \,\forall \bar{y} \,\exists \bar{z} \,\forall \bar{u}(\bar{u} \in \bar{z} \iff \bar{u} = \bar{x} \lor \bar{u} = \bar{y}). \\ &\textbf{Proof. } &&\text{Let } \bar{z} = \{\bar{x}, \bar{y}\}, \text{ and use absoluteness of pairing operation.} \\\\ \end{alignat}$$

$$\begin{alignat}{2} &B1_l: &&\exists \bar{E} \,\forall \bar{u} \,\forall \bar{v}((\bar{u}, \bar{v}) \in \bar{E} \iff \bar{u} \in \bar{v}). \\ &\textbf{Proof. } \;\;&&\text{Using B1, there exists an } E: \forall u \forall v((u, v) \in E \iff u \in v). \\ & &&\text{Let } \bar{E} = E \cap L. \text{ By a previous theorem: } \mathfrak{L}(E \cap L). \\ & &&(\bar{u}, \bar{v}) \in \bar{E} \implies (\bar{u}, \bar{v}) \in E \implies \bar{u} \in \bar{v}. \\ & &&\bar{u} \in \bar{v} \implies (\bar{u}, \bar{v}) \in E. \;\text{ Also, } (\bar{u}, \bar{v}) \in L. \\ & &&\text{Thus, } \;\bar{u} \in \bar{v} \implies (\bar{u}, \bar{v}) \in E \cap L = \bar{E}. \\ & &&\therefore \;\,\forall \bar{u} \,\forall \bar{v}((\bar{u}, \bar{v}) \in \bar{E} \iff \bar{u} \in \bar{v}). \\\\

&B2_l: &&\forall \bar{A} \,\forall \bar{B} \,\exists \bar{C} \,\forall \bar{u}(\bar{u} \in \bar{C} \iff \bar{u} \in \bar{A} \land \bar{u} \in \bar{B}). \\ &\textbf{Proof. } &&\text{Using B2, let } \bar{C} = \bar{A} \cap \bar{B}. \;\text{ By a previous theorem: } \mathfrak{L}(\bar{A} \cap \bar{B}).\\ & &&\text{B2 implies } \,\forall u(u \in \bar{C} \iff u \in \bar{A} \land u \in \bar{B}). \\ & &&\therefore \;\forall \bar{u}(\bar{u} \in \bar{C} \iff \bar{u} \in \bar{A} \land \bar{u} \in \bar{B}). \\ & &&\text{We can also prove that } \cap_l \text{ is absolute, but this is not needed here.} \\\\

&B3_l: &&\forall \bar{A} \,\exists \bar{B} \,\forall \bar{u}(\bar{u} \in \bar{B} \iff \bar{u} \notin \bar{A}). \\ &\textbf{Proof. } &&\text{Let } \bar{B} = L \setminus \bar{A}. \;\text{ By a previous theorem: } \mathfrak{L}(L \setminus \bar{A}).\\ & &&\forall u(u \in \bar{B} \iff u \in L \land u \notin \bar{A}). \\ & &&\therefore \;\bar{u}(\bar{u} \in \bar{B} \iff \bar{u} \in L \land \bar{u} \notin \bar{A} \iff \bar{u} \notin \bar{A}). \\\\

&B4_l: &&\forall \bar{A} \,\exists \bar{B} \,\forall \bar{u}(\bar{u} \in \bar{B} \iff \exists \bar{v}((\bar{v}, \bar{u}) \in \bar{A}). \\ &\textbf{Proof. } &&\text{Let } \bar{B} = Dom(\bar{A}). \;\text{ By a previous theorem: } \mathfrak{L}(Dom(\bar{A})).\\ & &&\forall u\bigl[u \in \bar{B} \iff  \exists v((v, u) \in \bar{A})\bigr]. \\ & &&\therefore \;\forall \bar{u}\bigl[\bar{u} \in \bar{B} \iff \exists v((v, \bar{u}) \in \bar{A}) \iff \exists \bar{v}((\bar{v}, \bar{u}) \in \bar{A})\bigr].  \\\\

&B5_l: &&\forall \bar{A} \,\exists \bar{B} \,\forall \bar{u}\, \forall \bar{v}((\bar{v}, \bar{u}) \in \bar{B} \iff \bar{u} \in \bar{A}). \\ &\textbf{Proof. } &&\text{Let } \bar{B} = L \times \bar{A}. \;\text{ By a previous theorem: } \mathfrak{L}(L \times \bar{A}).\\ & &&\forall u \,\forall v\bigl[(v,u) \in \bar{B} \iff (\bar{v} \in L \land \bar{u} \in \bar{A})\,\bigr]. \\ & &&\therefore \;\forall \bar{u} \,\forall \bar{v}\bigl[(\bar{v}, \bar{u}) \in \bar{B} \iff (\bar{v} \in L \land \bar{u} \in \bar{A}) \iff \bar{u} \in \bar{A}\,\bigr]. \\\\

&B6_l: &&\forall \bar{A} \,\exists \bar{B} \,\forall \bar{u} \,\forall \bar{v} \,\forall \bar{w}((\bar{u},\bar{v},\bar{w}) \in \bar{B} \iff (\bar{v},\bar{u},\bar{w}) \in \bar{A}). \\ &\textbf{Proof. } &&\text{Let } \bar{B} = Tr(\bar{A}). \;\text{ By a previous theorem: } \mathfrak{L}(Tr(\bar{A})).\\ & &&\forall u \,\forall u \,\forall w \bigl[(u,v,w) \in \bar{B} \iff (v,u,w) \in \bar{A}\bigr]. \\ & &&\therefore \;\forall \bar{u} \,\forall \bar{v} \,\forall \bar{w}\bigl[(\bar{u},\bar{v},\bar{w}) \in \bar{B} \iff (\bar{v},\bar{u},\bar{w}) \in \bar{A}\bigr]. \\\\

&B7_l: &&\forall \bar{A} \,\exists \bar{B} \,\forall \bar{u} \,\forall \bar{v} \,\forall \bar{w}((\bar{w},\bar{u},\bar{v}) \in \bar{B} \iff (\bar{u},\bar{v},\bar{w}) \in \bar{A}). \\ &\textbf{Proof. } &&\text{Let } \bar{B} = Cp(\bar{A}). \;\text{ By a previous theorem: } \mathfrak{L}(Cp(\bar{A})).\\ & &&\forall u \,\forall u \,\forall w \bigl[(w,u,v) \in \bar{B} \iff (u,v,w) \in \bar{A}\bigr]. \\ & &&\therefore \;\forall \bar{u} \,\forall \bar{v} \,\forall \bar{w}\bigl[(\bar{w},\bar{u},\bar{v}) \in \bar{B} \iff (\bar{u},\bar{v},\bar{w}) \in \bar{A}\bigr]. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &C1_l: &&\exists \bar{a}\bigl[\neg \mathfrak{Em}(\bar{a}) \land \forall \bar{x}(\bar{x} \in \bar{a} \implies \exists \bar{y}(\bar{y} \in \bar{a} \land \bar{x} \subset \bar{y}))\bigr]. \\ &\textbf{Proof. } \;\;&&\text{Let } \bar{a} = F(J(0, \omega, 0)). \\ & &&(i, \alpha, \beta) \;S\; (0, \omega, 0) \implies (\alpha, \beta) \;R\; (\omega, 0) \lor ((\alpha, \beta) = (\omega, 0) \land i < 0 \text{ [latter disjunct cannot occur.]} \\ & &&(\alpha, \beta) \;R\; (\omega, 0) \iff Max(\alpha, \beta) < Max(\omega, 0) \lor (Max(\alpha, \beta) = Max(\omega, 0) \land (\alpha, \beta) \;Le\; (\omega, 0).) \\ & &&(\alpha, \beta) \;Le\; (\omega, 0) \iff \beta < 0 \lor (\beta = 0 \land \alpha < \omega). \\ & &&\text{Last two cases imply: } \alpha, \beta < \omega. \\ & &&\therefore \;\bar{x} \in \bar{a} \implies \bar{x} = F(J(i, \alpha, \beta)) < F(J(0, \omega, 0)) \implies \alpha, \beta < \omega \implies J(i, \alpha, \beta) = n < \omega. \\ & &&\empty \in \bar{a} \implies \neg \mathfrak{Em}(\bar{a}). \\ & &&\text{We now prove } \bar{x} \in \bar{a} \implies \bar{x} \cup \{\bar{x}\} \in \bar{a}:  \;\,\bar{x} \in \bar{a} \implies \exists n < \omega (\bar{x} = F(n)). \\ & &&\text{Then } \{\bar{x}\} = \{\bar{x},\bar{x}\} = F(J((p, n, n)) = F(J(n_1)). \;\;\bar{x} \cup \{\bar{x}\} \subseteq \bar{z} = F(J((0, n_1+1, 0))). \\ & &&\text{Then } \bar{x} \cup \{\bar{x}\} = \bar{z} \setminus (z \setminus \bar{x}) \setminus \{\bar{x}\}) = F(n_2) \in F(J((0, \omega, 0))). \\ & &&\text{Let } \bar{y} = \bar{x} \cup \{\bar{x}\}. \;\;\therefore \;\bar{y} \in \bar{a} \land \bar{x} \subset \bar{y}. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &\textbf{Definition. } \;\;(\alpha, \beta) \;Le\; (\gamma, \delta) \iff \beta < \delta \lor (\beta = \delta \land \alpha < \gamma). \\ &\textbf{Definition. } \;\;(\alpha, \beta) \;R\; (\gamma, \delta) \iff Max(\alpha, \beta) < Max(\gamma, \delta) \lor (Max(\alpha, \beta) = Max(\gamma, \delta) \land (\alpha, \beta) \;Le\; (\gamma, \delta)) \\ &\textbf{Definition. } \;\;(i, \alpha, \beta) \;S\; (j, \gamma, \delta) \iff (\alpha, \beta) \;R\; (\gamma, \delta) \lor ((\alpha, \beta) = (\gamma, \delta) \land i < j. \\\\ \end{alignat}$$

$$\begin{alignat}{2} &C2_l: &&\forall \bar{x} \,\exists \bar{y} \,\forall \bar{u} \,\forall \bar{v} \bigl[(\bar{u} \in \bar{v} \land \bar{v} \in \bar{x}) \implies \bar{u} \in \bar{y}\bigr]. \\ &\textbf{Proof. } \;\;&&\text{Using axiom C2: } \text{ Let } y \text{ satisfy } \forall u \,\forall v(u \in v \land v \in \bar{x} \implies u \in y). \\ & &&\text{Since } L \cap y \subseteq L, \text{ there is a } \bar{z} \text{ such that } L \cap y \subseteq \bar{z}. \\ & &&\bar{z} \text{ satisfies } C2_l: (\bar{u} \in \bar{v} \land \bar{v} \in \bar{x}) \implies \bar{u} \in y \implies \bar{u} \in L \cap y \implies \bar{u} \in \bar{z}. \\\\

&C3_l: &&\forall \bar{x} \,\exists \bar{y} \,\forall \bar{u} \bigl[(\bar{u} \subseteq \bar{x} \implies \bar{u} \in \bar{y}\bigr]. \\ &\textbf{Proof. } &&\text{Using axiom C3: } \text{ Let } y \text{ satisfy } \forall u(u \subseteq \bar{x} \implies u \in y). \\ & &&\text{Since } L \cap y \subseteq L, \text{ there is a } \bar{z} \text{ such that } L \cap y \subseteq \bar{z}. \\ & &&\bar{z} \text{ satisfies } C3_l: \bar{u} \subseteq \bar{x} \implies \bar{u} \in y \implies \bar{u} \in L \cap y \implies \bar{u} \in \bar{z}. \\\\

&C4_l: &&\forall \bar{x} \,\forall \bar{F} \bigl[Un(\bar{F}) \implies \exists \bar{y} \,\forall \bar{v}(\bar{v} \in \bar{y} \iff \exists \bar{u}(\bar{u} \in \bar{x}) \land (\bar{v}, \bar{u}) \in \bar{F})\bigr]. \\ &\textbf{Proof. } &&\text{Using axiom C4: Given } \bar{x} \text{ and } \bar{F}, \text{ there is a } y \text{ such that } \forall v(v \in y \iff \exists u(u \in \bar{x} \land (v, u) \in \bar{F})). \\ & &&\text{Now } y = \bar{F}[\bar{x}] \implies \mathfrak{M}(\bar{F}[\bar{x}]). \;\text{By a previous theorem, } \mathfrak{L}(\bar{F}[\bar{x}]). \;\,\therefore y \in L. \\ & &&\text{Since } (v \in y \implies v \in L) \land (u \in \bar{x} \implies u \in L), \text{ we have: } \,\exists \bar{y} \,\forall \bar{v}(\bar{v} \in \bar{y} \iff \exists \bar{u}(\bar{u} \in \bar{x}) \land (\bar{v}, \bar{u}) \in \bar{F}). \\\\

& && \text{*********** CHECK NEXT 2: HAVE WE STATED D1 and E1 correctly?? } \\\\

&D1_l: &&\neg \mathfrak{Em}(\bar{A}) \implies \exists \bar{u}\bigl[\bar{u} \in \bar{A} \land \mathfrak{Ex}(\bar{u}, \bar{A})\bigr]. \\ &\textbf{Proof. } &&\text{Let } \neg \mathfrak{Em}(\bar{A}). \;\text{Using axiom D1: } \text{ Let } u \text{ satisfy } u \in \bar{A} \land \mathfrak{Ex}(u, \bar{A})). \\ & &&\text{Since } u \in \bar{A} \implies u \in L, \text{ we have: } \,\exists \bar{u}\bigl[\bar{u} \in \bar{A} \land \mathfrak{Ex}(\bar{u}, \bar{A}\bigr]. \\\\

&E1_l: &&\exists \bar{G} \,\forall \bar{x}\bigl[\mathfrak{Un}(\bar{G}) \land \exists \bar{y}(\neg \mathfrak{Em}(\bar{x}) \implies (\bar{y}, \bar{x}) \in \bar{G})\bigr]. \\ &\textbf{Proof. } &&\text{Define: } (\bar{y}, \bar{x}) \in \bar{G} \iff \bar{y} \in \bar{x} \land \forall \bar{z}(Od(\bar{y}) \le Od(\bar{z})). \\ & &&\text{Must prove: } \mathfrak{L}(\bar{G}). \text{Follows from Class Existence Theorem, but must prove that Od is absolute.} \\ \end{alignat}$$