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Proposition 8.12. \( \\left( {\\alpha + \\beta }\\right) + \\gamma = \\alpha + \\left( {\\beta + \\gamma }\\right) \) . | Proof (By transfinite induction on \( \\gamma \) ). For \( \\gamma = 0 \) we note that \( \\left( {\\alpha + \\beta }\\right) + 0 = \) \( \\alpha + \\beta = \\alpha + \\left( {\\beta + 0}\\right) \) . If \( \\left( {\\alpha + \\beta }\\right) + \\gamma = \\alpha + \\left( {\\beta + \\gamma }\\right) \) then \( \\left( ... | Yes |
Proposition 8.13. \( \alpha \geqq \omega \rightarrow \left( {\exists !\beta }\right) \left( {\exists !n}\right) \left\lbrack {\beta \in {K}_{\text{II }} \land \alpha = \beta + n}\right\rbrack \) | Proof. If \( A = \left\{ {\gamma \in {K}_{\mathrm{{II}}} \mid \gamma \leqq \alpha }\right\} \) and if \( \beta = \cup \left( A\right) \), then \( \beta \in {K}_{\mathrm{{II}}} \) and \( \beta \leqq \alpha \) . Therefore, by Proposition 8.8, \( \left( {\exists \gamma }\right) \left\lbrack {\beta + \gamma = \alpha }\righ... | Yes |
(1) \( 0 \cdot \alpha = \alpha \cdot 0 = 0 \) . | Proof (By transfinite induction). (1) By definition \( \alpha \cdot 0 = 0 \) for all \( \alpha \) including \( \alpha = 0 \) . If \( 0 \cdot \alpha = 0 \) then \( 0\left( {\alpha + 1}\right) = 0 \cdot \alpha + 0 = 0 \) . If \( \alpha \in {K}_{\text{II }} \) and \( 0 \cdot \gamma = 0 \) for \( \gamma < \alpha \), then\n... | No |
Proposition 8.24. \( \alpha \neq 0 \land \beta \in {K}_{\mathrm{{II}}} \rightarrow {\alpha \beta } \in {K}_{\mathrm{{II}}} \) . | Proof. If \( \alpha \neq 0 \land \beta \in {K}_{\text{II }} \) then \( {\alpha \beta } \neq 0 \) . Therefore \( {\alpha \beta } \in {K}_{\text{II }} \) or \( \left( {\exists \gamma }\right) \lbrack \gamma + 1 \) \( = {\alpha \beta }\rbrack \) . Since \( \gamma \in \gamma + 1 \) and since \( \beta \in {K}_{\mathrm{{II}}... | Yes |
Proposition 8.25. \( \\alpha \\left( {\\beta + \\gamma }\\right) = {\\alpha \\beta } + {\\alpha \\gamma } \). | Proof (By transfinite induction on \( \\gamma \) ). For \( \\gamma = 0 \) we see that \( \\alpha \\left( {\\beta + 0}\\right) = \) \( {\\alpha \\beta } = {\\alpha \\beta } + \\alpha \\cdot 0 \). If \( \\alpha \\left( {\\beta + \\gamma }\\right) = {\\alpha \\beta } + {\\alpha \\gamma } \), then \( \\alpha \\left( {\\bet... | Yes |
Proposition 8.27. \( \beta \neq 0 \rightarrow \left( {\exists !\gamma }\right) \left( {\exists !\delta }\right) \left\lbrack {\alpha = {\beta \gamma } + \delta \land \delta < \beta }\right\rbrack \) . | Proof. If \( \alpha < \beta \), then \( \alpha = \beta \cdot 0 + \alpha \land \alpha < \beta \) . If \( \beta \leqq \alpha \) and \( \gamma = \sup \{ \delta \mid {\beta \delta } \leqq \alpha \} \) then \( \gamma \geqq 1 \) . Suppose that \( \alpha < {\beta v} \) . Then \( {\beta \delta } \leqq \alpha \) implies \( \del... | Yes |
Corollary 8.28. \( n \neq 0 \rightarrow \left( {\exists !q}\right) \left( {\exists !r}\right) \left\lbrack {m = {nq} + r \land r < n}\right\rbrack \) | Proof. By Proposition 8.27 \( \left( {\exists !\gamma }\right) \left( {\exists !\delta }\right) \left\lbrack {m = {n\gamma } + \delta \land \delta < n}\right\rbrack \) . But \( {n\gamma } + \) \( \delta \in \omega \) implies \( {n\gamma } \in \omega \) and \( \delta \in \omega \) . Furthermore if \( 1 \leqq n \), then ... | No |
Proposition 8.29. \( \gamma \in {K}_{\text{II }} \land m \neq 0 \rightarrow m\left( {\gamma + n}\right) = \gamma + {mn} \) . | Proof (By induction on \( \gamma + n \) ). If \( \gamma + n = \omega \) we have \( {m\omega } = \mathop{\bigcup }\limits_{{n < \omega }}{mn} \) . Since \( {mn} < \omega \) we have\n\n\[ \mathop{\bigcup }\limits_{{n < \omega }}{mn} \leqq \omega \]\n\nFurthermore \( p < \omega \rightarrow \left( {\exists q}\right) \left(... | Yes |
(1) \( {0}^{0} = 1 \) .\n\n(2) \( {0}^{\beta } = 0,\beta \geqq 1 \) .\n\n(3) \( {1}^{\beta } = 1 \) . | Proof. (1) From Definition \( {8.30},{0}^{ \circ } = 1 \) .\n\n(2) If \( \beta \geqq 1 \), then \( \beta \in {K}_{\mathrm{{II}}} \) or \( \left( {\exists \delta }\right) \left\lbrack {\beta = \delta + 1}\right\rbrack \) . If \( \beta \in {K}_{\mathrm{{II}}} \) then by Definition \( {8.30},{0}^{\beta } = 0 \) . If \( \b... | Yes |
Proposition 8.32. \( 1 \leqq \alpha \rightarrow 1 \leqq {\alpha }^{\beta } \) . | Proof (By transfinite induction on \( \beta \) ). First we note that \( {\alpha }^{0} = 1 \) . If \( 1 \leqq {\alpha }^{\beta } \) then since \( 1 \leqq \alpha \) we have \( 1 \leqq {\alpha }^{\beta } \leqq {\alpha }^{\beta } \cdot \alpha \), i.e., \( 1 \leqq {\alpha }^{\beta + 1} \) . If \( \beta \in {K}_{\text{II }} ... | Yes |
Proposition 8.33. \( \alpha < \beta \land 1 < \gamma \rightarrow {\gamma }^{\alpha } < {\gamma }^{\beta } \) . | Proof (By transfinite induction on \( \beta \) ). If \( \beta = \alpha + 1 \), then \( 1 < \gamma \) implies \( {\gamma }^{\alpha } < {\gamma }^{\alpha + 1} \) . Suppose that\n\n\[ \alpha < \beta \land 1 < \gamma \rightarrow {\gamma }^{\alpha } < {\gamma }^{\beta } \]\n\nIf \( \alpha < \beta + 1 \), then \( \alpha < \b... | Yes |
Corollary 8.34. \( 1 < \gamma \land {\gamma }^{\alpha } < {\gamma }^{\beta } \rightarrow \alpha < \beta \) . | Proof. By Proposition 8.33, \( \beta \leqq \alpha \land 1 < \gamma \rightarrow {\gamma }^{\beta } \leqq {\gamma }^{\alpha } \) . | Yes |
Proposition 8.35. \( \alpha < \beta \rightarrow {\alpha }^{\gamma } \leqq {\beta }^{\gamma } \) . | Proof (By transfinite induction on \( \gamma \) ). For \( \gamma = 0 \) we have \( {\alpha }^{0} = 1 = {\beta }^{0} \). Suppose that \( \alpha < \beta \) and \( {\alpha }^{\gamma } \leqq {\beta }^{\gamma } \). Then \( {\alpha }^{\gamma + 1} = {\alpha }^{\gamma } \cdot \alpha \leqq {\beta }^{\gamma } \cdot \alpha < {\be... | Yes |
Corollary 8.36. \( \alpha < \beta \land \gamma \in {K}_{\mathrm{I}} \land \gamma \neq 0 \rightarrow {\alpha }^{\gamma } < {\beta }^{\gamma } \) . | Proof. If \( \gamma \in {K}_{1} \land \gamma \neq 0 \), then \( \left( {\exists \delta }\right) \left\lbrack {\gamma = \delta + 1}\right\rbrack \) . By Proposition 8.35, if \( \alpha < \beta \), then \( {\alpha }^{\delta } \leqq {\beta }^{\delta } \) . But \( {\alpha }^{\gamma } = {\alpha }^{\delta } \cdot \alpha \leqq... | Yes |
Proposition 8.37. \( \alpha > 1 \rightarrow \beta \leqq {\alpha }^{\beta } \) . | Proof (By transfinite induction on \( \beta \) ). For \( \beta = 0 \) we have \( 0 \leqq {\alpha }^{0} = 1 \) . If \( \beta \leqq {\alpha }^{\beta } \) then \( \beta + 1 \leqq {\alpha }^{\beta } + 1 \) . But since \( \beta < \beta + 1 \) we have from Proposition 8.33 that \( {\alpha }^{\beta } < {\alpha }^{\beta + 1} \... | Yes |
Proposition 8.38. \( \alpha > 1 \land \beta > 0 \rightarrow \left( {\exists !\delta }\right) \left\lbrack {{\alpha }^{\delta } \leqq \beta < {\alpha }^{\delta + 1}}\right\rbrack \) . | Proof. Since by Proposition \( {8.37},\beta \leqq {\alpha }^{\beta } \) and since \( {\alpha }^{\beta } < {\alpha }^{\beta + 1} \) there exists a smallest ordinal \( \gamma \) such that \( \beta < {\alpha }^{\gamma } \) . From Definition 8.30 it follows that \( \gamma \in {K}_{\mathrm{I}} \) . Since \( {\alpha }^{0} = ... | Yes |
(1) \( \alpha > 1 \land \beta \in {K}_{\mathrm{{II}}} \rightarrow {\alpha }^{\beta } \in {K}_{\mathrm{{II}}} \) . | Proof. (1) If \( \alpha > 1 \), then \( {\alpha }^{\beta } \geqq 1 \) and hence \( {\alpha }^{\beta } \neq 0 \) . Therefore \( {\alpha }^{\beta } \in {K}_{\text{II }} \) or \( \left( {\exists \delta }\right) \left\lbrack {\delta + 1 = {\alpha }^{\beta }}\right\rbrack \) . Since \( \beta \in {K}_{\text{II }} \) and \( \... | No |
Proposition 8.41. \( {\alpha }^{\beta } \cdot {\alpha }^{\gamma } = {\alpha }^{\beta + \gamma } \) . | Proof (By transfinite induction on \( \gamma \) ). First we note that \( {\alpha }^{\beta } \cdot {\alpha }^{0} = {\alpha }^{\beta } \cdot 1 = \) \( {\alpha }^{\beta } = {\alpha }^{\beta + 0} \) . If \( {\alpha }^{\beta } \cdot {\alpha }^{\gamma } = {\alpha }^{\beta + \gamma } \) then \( {\alpha }^{\beta } \cdot {\alph... | Yes |
Proposition 8.42. \( {\left( {\alpha }^{\beta }\right) }^{\gamma } = {\alpha }^{\beta \gamma } \) . | Proof (By transfinite induction on \( \gamma \) ). For \( \gamma = 0 \) we have \( {\left( {\alpha }^{\beta }\right) }^{0} = 1 = {\alpha }^{\beta \cdot 0} \) . If \( {\left( {\alpha }^{\beta }\right) }^{\gamma } = {\alpha }^{\beta \gamma } \) then \( {\left( {\alpha }^{\beta }\right) }^{\gamma + 1} = {\left( {\alpha }^... | Yes |
Proposition 8.43.\n\n\[ \alpha > 1 \land {\gamma }_{n} < \alpha \land \cdots \land {\gamma }_{0} < \alpha \land 0 \leqq {\beta }_{0} < \cdots < {\beta }_{n} < \beta \]\n\n\[ \rightarrow {\alpha }^{{\beta }_{n}}{\gamma }_{n} + \cdots + {\alpha }^{{\beta }_{0}}{\gamma }_{0} < {\alpha }^{\beta }. \] | Proof (By induction on \( n \) ). If \( n = 0 \) then since \( {\gamma }_{0} < \alpha \) we have that \( {\alpha }^{{\beta }_{0}}{\gamma }_{0} < \) \( {\alpha }^{{\beta }_{0} + 1} \leqq {\alpha }^{\beta } \) . If \( n > 0 \) then since \( {\beta }_{n - 1} < {\beta }_{n} < \beta \) we have as our induction hypothesis\n\... | Yes |
Proposition 8.44. \( \\beta > 0 \\land \\alpha > 1 \\rightarrow \\left( {\\exists !n}\\right) \\left( {\\exists !{\\beta }_{0}}\\right) \\cdots \\left( {\\exists !{\\beta }_{n}}\\right) \\left( {\\exists !{\\gamma }_{0}}\\right) \\cdots \\left( {\\exists !{\\gamma }_{n}}\\right) \) \( \\lbrack \\beta = {\\alpha }^{{\\b... | Proof (By transfinite induction on \( \\beta \) ). By Proposition 8.38, there exists a \( \\delta \) such that\n\n(1) \( {\\alpha }^{\\delta } < \\beta < {\\alpha }^{\\delta + 1} \).\n\nBy Proposition 8.27 there exists a \( \\tau \) and \( v \) such that\n\n(2) \( \\beta = {\\alpha }^{\\delta }\\tau + v \)\n\nand \( v ... | No |
Proposition 8.45. \( \alpha > 1 \land {\beta }_{0} < {\beta }_{1} < \cdots < {\beta }_{n} \land 0 < {\gamma }_{0} < \alpha \land \cdots \land 0 < {\gamma }_{n} < \alpha \land \delta \geqq \omega \rightarrow \left( {{\alpha }^{{\beta }_{n}}{\gamma }_{n} + \cdots + {\alpha }^{{\beta }_{0}}{\gamma }_{0}}\right) {\alpha }^... | Proof. From Proposition 8.43.\n\n\[ \n{\alpha }^{{\beta }_{n}} \leqq {\alpha }^{{\beta }_{n}}{\gamma }_{n} + \cdots + {\alpha }^{{\beta }_{0}}{\gamma }_{0} < {\alpha }^{{\beta }_{n} + 1}. \n\]\n\nTherefore\n\n\[ \n{\alpha }^{{\beta }_{n} + \delta } = {\alpha }^{{\beta }_{n}}{\alpha }^{\delta } \leqq \left( {{\alpha }^{... | Yes |
Proposition 8.46. \( \;\alpha \in {K}_{\mathrm{{II}}} \land {\beta }_{0} < {\beta }_{1} < \cdots < {\beta }_{n} \land 0 < {m}_{n} \land \delta > 0 \rightarrow \left( {{\alpha }^{{\beta }_{n}}{m}_{n}}\right. \) \( \left. {+\cdots + {\alpha }^{{\beta }_{0}}{m}_{0}}\right) {\alpha }^{\delta } = {\alpha }^{{\beta }_{n} + \... | Proof. From Proposition 8.43.\n\n\[ \n{\alpha }^{{\beta }_{n - 1}}{m}_{n - 1} + \cdots + {\alpha }^{{\beta }_{0}}{m}_{0} < {\alpha }^{{\beta }_{n}}. \n\] \n\nTherefore \n\n\[ \n{\alpha }^{{\beta }_{n}} \leqq {\alpha }^{{\beta }_{n}}{m}_{n} + \cdots + {\alpha }^{{\beta }_{0}}{m}_{0} < {\alpha }^{{\beta }_{n}}{m}_{n} + {... | Yes |
Proposition 8.47. If \( \alpha \in {K}_{\text{II }} \land \beta > 0 \land m > 0 \) then\n\n(1) \( {\left( {\alpha }^{\beta }m\right) }^{\gamma } = {\alpha }^{\beta \gamma }m,\gamma \in {K}_{\mathrm{I}} \land \gamma \neq 0 \) .\n\n(2) \( {\left( {\alpha }^{\beta }m\right) }^{\gamma } = {\alpha }^{\beta \gamma },\gamma \... | Proof (By transfinite induction on \( \gamma \) ). If \( \gamma = 1 \), then \( {\left( {\alpha }^{\beta }m\right) }^{1} = {\alpha }^{\beta \cdot 1}m \) . If \( {\left( {\alpha }^{\beta }m\right) }^{\gamma } = {\alpha }^{\beta \gamma }m \) then \( {\left( {\alpha }^{\beta }m\right) }^{\gamma + 1} = {\left( {\alpha }^{\... | Yes |
Proposition 8.48. \[ \alpha \in {K}_{\mathrm{{II}}} \land {\beta }_{0} < {\beta }_{1} < \cdots < {\beta }_{n} \rightarrow {\left( {\alpha }^{{\beta }_{n}}{m}_{n} + \cdots + {\alpha }^{{\beta }_{0}}{m}_{0}\right) }^{\gamma } \leqq {\alpha }^{{\beta }_{n}\gamma }\left( {{m}_{n} + 1}\right) . \] | Proof. Note that \( {\alpha }^{{\beta }_{n}}{m}_{n} + \cdots + {\alpha }^{{\beta }_{0}}{m}_{0} \leqq {\alpha }^{{\beta }_{n}}\left( {{m}_{n} + 1}\right) \) . Therefore by Proposi- tion \( {8.47}{\left( {\alpha }^{{\beta }_{n}}{m}_{n} + \cdots + {\alpha }^{{\beta }_{0}}{m}_{0}\right) }^{\gamma } \leqq {\left\lbrack {\al... | Yes |
\[ \alpha \in {K}_{\mathrm{{II}}} \land {\beta }_{0} < {\beta }_{1} < \cdots < {\beta }_{n} \land \gamma \in {K}_{\mathrm{{II}}} \rightarrow {\left( {\alpha }^{{\beta }_{n}}{m}_{n} + \cdots + {\alpha }^{{\beta }_{0}}{m}_{0}\right) }^{\gamma } = {\alpha }^{{\beta }_{n}\gamma }. \] | Proof. Note that \( {\alpha }^{{\beta }_{n}} \leqq {\alpha }^{{\beta }_{n}}{m}_{n} + \cdots + {\alpha }^{{\beta }_{0}}{m}_{0} \leqq {\alpha }^{{\beta }_{n}}\left( {{m}_{n} + 1}\right) \) . Therefore by Proposition 8.47 | No |
Corollary 8.50.\n\n\[ \alpha \in {K}_{\mathrm{{II}}} \land {\beta }_{0} < {\beta }_{1} < \cdots < {\beta }_{n} \land \gamma > 0 \rightarrow {\left( {\alpha }^{{\beta }_{n}}{m}_{n} + \cdots + {\alpha }^{{\beta }_{0}}{m}_{0}\right) }^{{\alpha }^{\gamma }} = {\alpha }^{{\beta }_{n}{\alpha }^{\gamma }}. \] | Proof. By Proposition \( {8.39},\gamma > 0 \land \alpha \in {K}_{\mathrm{{II}}} \rightarrow {\alpha }^{\gamma } \in {K}_{\mathrm{{II}}} \) . The result then follows from Proposition 8.49. | Yes |
Proposition 9.1. \( \left( {\forall x}\right) \left( {\exists y}\right) \left\lbrack {x \subseteq y \land \operatorname{Tr}\left( y\right) \land \left( {\forall z}\right) \left\lbrack {x \subseteq z \land \operatorname{Tr}\left( z\right) \rightarrow y \subseteq z}\right\rbrack }\right\rbrack \) . | Proof. If \( {G}^{c}x = x \cup \left( {\cup \left( x\right) }\right) \), then there exists a function \( f \) defined by recursion on \( \omega \) such that\n\n\[ \n{f}^{c}0 = x \n\]\n\n\[ \n{f}^{c}\left( {n + 1}\right) = {G}^{c}{f}^{c}n. \n\]\n\nFurthermore, if\n\n\[ \ny = \mathop{\bigcup }\limits_{{n < \omega }}{f}^{... | Yes |
Proposition 9.3. If \( R \) Wfr \( A \) and \( a \subseteq A \) then there exists a set \( b \) such that \( \left\lbrack {a \subseteq b \subseteq A}\right\rbrack \) and\n\n(1) \( \left( {\forall x \in A}\right) \left( {\forall y}\right) \left\lbrack {{xRy} \land y \in b \rightarrow x \in b}\right\rbrack \) .\n\n(2) \(... | Proof. (1) Since \( R \) is well founded on \( \mathrm{A} \), it follows, for each \( x \) in \( A \), that \( A \cap \) \( {\left( {R}^{-1}\right) }^{a}\{ x\} \) is a set. Therefore if\n\n\[ B = \left\{ {\left\langle {x, A \cap {\left( {R}^{-1}\right) }^{a}\{ x\} }\right\rangle \mid x \in A}\right\} \]\n\nthen \( B \)... | No |
Proposition 9.4. \[ R\operatorname{Wfr}A \land B \subseteq A \land B \neq 0 \rightarrow \left( {\exists x \in B}\right) \left\lbrack {B \cap {\left( {R}^{-1}\right) }^{\prime \prime }\{ x\} = 0}\right\rbrack . \] | Proof. Let \( a \in B \) . By Proposition 9.3 there exists a set \( b \) such that \[ \{ a\} \subseteq b \subseteq A \land \left( {\forall x \in A}\right) \left( {\forall y}\right) \left\lbrack {{xRy} \land y \in b \rightarrow x \in b}\right\rbrack . \] Then \( b \cap B \) is a nonempty subset of \( A \) . Therefore \[... | Yes |
Proposition 9.6. If \( \mathrm{{Cl}}\left( {{R}_{1}, A}\right) \land \cdots \land \mathrm{{Cl}}\left( {{R}_{p}, A}\right) \land {\mathrm{{Cl}}}_{2}\left( {{S}_{1}, A}\right) \land \cdots \land {\mathrm{{Cl}}}_{2}\left( {{S}_{q}, A}\right) \) if \( \left( {\forall x \subseteq A}\right) \left\lbrack {\mathcal{M}\left( {{... | Proof. If \( {G}^{c}x = x \cup {R}_{1}^{a}x \cup \cdots \cup {R}_{p}^{a}x \cup {S}_{1}^{a}{x}^{2} \cup \cdots \cup {S}_{q}^{a}{x}^{2}, x \subseteq A \) then there exists a function \( f \) defined on \( \omega \) by recursion such that\n\n\[ {f}^{c}0 = a \land {f}^{c}\left( {n + 1}\right) = {G}^{c}{f}^{c}n \]\n\nLet\n\... | Yes |
Proposition 9.12. \( \\left( {\\forall x \\in a}\\right) \\left\\lbrack {\\mathrm{{Wf}}\\left( x\\right) }\\right\\rbrack \\rightarrow \\mathrm{{Wf}}\\left( a\\right) \) . | Proof. If each element of \( a \) is well founded and \( x \\in a \), then \( \\left( {\\exists \\alpha }\\right) \\left\\lbrack {x \\in {R}_{1}^{c}\\alpha }\\right\\rbrack \) . If\n\n\[ \n{F}^{i}x = {\\mu }_{\\alpha }\\left( {x \\in {R}_{1}^{i}\\alpha }\\right) \n\]\n\nthen since \( F \) is a function \( {F}^{\\alpha ... | Yes |
Proposition 9.13. \( \mathrm{{Wf}}\left( a\right) \) . | Proof. From the Axiom of Regularity \( E \) is a well-founded relation on \( V \) . From Proposition 9.12 if\n\n\[ A = \{ x \mid \mathrm{{Wf}}\left( x\right) \} \]\n\nthen \( a \subseteq A \) implies \( a \in A \) . Then by \( \in \) -induction (Proposition 5.22) \( A = V \) . | Yes |
(1) \( \operatorname{rank}\left( a\right) \in \mathrm{{On}} \) . | Proof. (1) Definition 9.14. | No |
Proposition 9.16. \( a \in b \rightarrow \operatorname{rank}\left( a\right) < \operatorname{rank}\left( b\right) \) | Proof. By Proposition 9.15, if \( \alpha = \operatorname{rank}\left( a\right) \), then \( a \notin {R}^{c}\alpha \) . If \( a \in b \) then since \( a \notin {R}_{1}^{c}\alpha \) it follows that \( b \nsubseteq {R}_{1}^{c}\alpha \) and hence \( b \notin {R}_{1}^{c}\left( {\alpha + 1}\right) \) . Thus \( \alpha < \opera... | Yes |
Proposition 9.17. \( \operatorname{rank}\left( a\right) = {\mu }_{\beta }\left( {\left( {\forall x \in a}\right) \left\lbrack {\operatorname{rank}\left( x\right) < \beta }\right\rbrack }\right) \) . | Proof. If \( x \in a \), then, by Proposition 9.15,\n\n\[ \operatorname{rank}\left( x\right) < \operatorname{rank}\left( a\right) .\n\]\n\nIf \( x \in a \) and in addition, \( \operatorname{rank}\left( x\right) < \beta \), then\n\n\[ x \in {R}_{1}^{i}\left( {\operatorname{rank}\left( x\right) + 1}\right) \leqq {R}_{1}^... | Yes |
Proposition 9.19. \( \\left( {\\exists \\alpha }\\right) \\left( {\\forall x \\in A}\\right) \\left\\lbrack {\\operatorname{rank}\\left( x\\right) \\leqq \\alpha }\\right\\rbrack \\rightarrow \\mathcal{M}\\left( A\\right) \) . | Proof. If \( \\operatorname{rank}\\left( x\\right) \\leqq \\alpha \), then \( x \\in {R}_{1}^{\\zeta }\\left( {\\alpha + 1}\\right) \) . Hence \( A \\subseteq {R}_{1}^{\\zeta }\\left( {\\alpha + 1}\\right) \) . | Yes |
Theorem 9.20 (Axiom 6’). \( A \neq 0 \rightarrow \left( {\exists x \in A}\right) \left\lbrack {x \cap A = 0}\right\rbrack \) . | Proof. If \( B = \{ \) rank \( \left( x\right) |x \in A\} \) and \( A \neq 0 \), then \( B \neq 0 \) . Thus \( B \) is a nonempty class of ordinals, hence, by Proposition 6.26, which was proved using only the weak form of the Axiom of Regularity, \( B \) has an E-minimal element \( \alpha \) . Since \( \alpha \in B \) ... | Yes |
Proposition 9.21. \( R\operatorname{Fr}A \land B \subseteqq A \land B \neq 0 \rightarrow \left( {\exists x \in B}\right) \left\lbrack {B \cap {\left( {R}^{-1}\right) }^{\alpha }\{ x\} = 0}\right\rbrack \) . | Proof (By contradiction). Suppose that\n\n\[ \n{B}_{0} = \{ x \in B \mid \left( {\forall y \in B}\right) \left\lbrack {\operatorname{rank}\left( x\right) \leqq \operatorname{rank}\left( y\right) }\right\rbrack \} \n\] \n\nand \n\n\[ \n{B}_{n + 1} = \left\{ {x \in B \mid \left( {\exists y \in {B}_{n}}\right) \left\lbrac... | Yes |
Proposition 9.22. If \( R\operatorname{Fr}A \land B \subseteq A \land \left( {\forall x \in A}\right) \left\lbrack {A \cap {\left( {R}^{-1}\right) }^{\alpha }\{ x\} \subseteq B \rightarrow }\right. \) \( x \in B\rbrack \) then \( A = B \) . | The proof is left to the reader. | No |
Proposition 10.9. \( \alpha \in N \leftrightarrow \alpha = \bar{\alpha } \) . | Proof. If \( \alpha = \bar{\alpha } \), then \( \alpha \in N \) by Definition 10.7. Conversely if \( \alpha \in N \), then \( \left( {\exists x}\right) \left\lbrack {\alpha = \bar{x}}\right\rbrack \) . Suppose that \( \bar{\alpha } < \alpha \), then since \( \bar{\alpha } \simeq \alpha \) and \( \alpha \simeq x \) we w... | Yes |
Proposition 10.12. \( a \subseteqq \alpha \rightarrow \left( {\exists \beta \leqq \alpha }\right) \left\lbrack {a \simeq \beta }\right\rbrack \) . | Proof. If \( a \subseteq \alpha \), then by Corollary 7.54, \( \left( {\exists \beta }\right) \left( {\exists f}\right) \left\lbrack {f{\operatorname{Isom}}_{E, E}\left( {\beta, a}\right) }\right\rbrack \) . This function \( f \) is then a strictly monotone ordinal function and so it follows that \( \left( {\forall \ga... | Yes |
Theorem 10.14 (Cantor-Schröder-Bernstein).\n\n\[ a \simeq c \leqq b \land b \simeq d \leqq a \rightarrow a \simeq b. \] | *Proof. If \( a \simeq c \leqq b \), then \( \bar{a} = \bar{c} \leqq \bar{b} \) . If \( b \simeq d \leqq a \), then \( \bar{b} = \bar{d} \leqq \bar{a} \) . Since \( \bar{a} = \bar{b} \) it follows, from *Proposition 10.10 that \( a \simeq b \) . | No |
Proposition 10.15. \( a \simeq b \rightarrow \mathcal{P}\left( a\right) \simeq \mathcal{P}\left( b\right) \) . | Proof. If \( a \simeq b \), then \( \left( {\exists f}\right) \left\lbrack {f : a\xrightarrow[\text{onto }]{1 - 1}b}\right\rbrack \) . Let \( F = \left\{ {\left\langle {x,{f}^{a}x}\right\rangle \mid x \in \mathcal{P}\left( a\right) }\right\} \) . Then \( F : \mathcal{P}\left( a\right) \rightarrow \mathcal{P}\left( b\ri... | Yes |
Theorem 10.19. For each set a\n\n\\[ \n\\left( {\\exists \\beta \\in N}\\right) \\left\\lbrack {\\beta = \\left\\{ {\\alpha \\mid \\left( {\\exists f}\\right) \\left\\lbrack {f : \\alpha \\overset{1 - 1}{ \\rightarrow }a}\\right\\rbrack }\\right\\} \\land \\neg \\left( {\\exists f}\\right) \\left\\lbrack {f : \\beta \\... | Proof. We consider all ordered pairs \\( \\langle r, x\\rangle \\) where \\( x \\subseteq a \\) and \\( r \\subseteq a \\times a \\) . If \\( r \\) well orders \\( x \\), then\n\n\\[ \n\\left( {\\exists !\\beta }\\right) \\left( {\\exists !{f}_{r, x}}\\right) \\left\\lbrack {{f}_{r, x}{\\operatorname{Isom}}_{r, E}\\lef... | Yes |
Proposition 10.20. \( m \simeq n \rightarrow m = n \) . | Proof (By induction on \( n \) ). If \( m \simeq 0 \), then \( m = 0 \) . As our induction hypothesis assume that \( \left( {\forall m}\right) \left\lbrack {m \simeq n \rightarrow m = n}\right\rbrack \) and assume that \( m \simeq \left( {n + 1}\right) \) . It then follows that \( m \neq 0 \) and hence, for some intege... | No |
(1) \( \neg \left( {n \simeq n + 1}\right) \) .\n\n(2) \( \neg \left( {\exists f}\right) \left\lbrack {f : \left( {n + 1}\right) \xrightarrow[]{1 - 1}n}\right\rbrack \) . | The proof is left to the reader. | No |
Proposition 10.22. \( \alpha \simeq n \rightarrow \alpha = n \) . | Proof. If \( \alpha \geqq \omega \) then \( n < \alpha \) and hence \( n + 1 \leqq \alpha \) . If \( \alpha \simeq n \) then, since \( n \subset n + 1 \), it follows from the Cantor-Schröder-Bernstein theorem that \( \alpha \simeq n + 1 \) . But then \( n \simeq n + 1 \) . This contradicts Corollary 10.21(1) and compel... | Yes |
Corollary 10.23. \( \bar{n} = n \) . | Proof. Proposition 10.24 and Definition 10.4. | No |
Corollary 10.24. \( \omega \subseteqq N \) . | Proof. Corollary 10.23 and Definition 10.7. | No |
Proposition 10.26. \( \Pr \left( {N}^{\prime }\right) \) . | Proof. Since \( N = {N}^{\prime } \cup \omega \) it follows that if \( {N}^{\prime } \) is a set, so also is \( N \) . | No |
Proposition 10.30. \( \operatorname{Fin}\left( a\right) \land b \subseteqq a \rightarrow \operatorname{Fin}\left( b\right) \) . | Proof. If \( a \) is finite then by Definition 10.29, \( \left( {\exists n}\right) \left\lbrack {a \simeq n}\right\rbrack \) . If \( b \subseteq a \) it then follows from Proposition 10.12 that \( \left( {\exists \beta \leqq n}\right) \left\lbrack {b \simeq \beta }\right\rbrack \) . Since such a \( \beta \) must be in ... | Yes |
Proposition 10.31. \( \operatorname{Fin}\left( a\right) \land \operatorname{Fin}\left( b\right) \rightarrow \operatorname{Fin}\left( {a \cup b}\right) \land \operatorname{Fin}\left( {a \times b}\right) \) | Proof (By induction on \( \bar{a} \) ). If \( \bar{a} = 0 \), then \( a = 0 \) and hence \( a \cup b = b \) and \( a \times b = 0 \) . But \( b \) is finite by hypothesis and 0 is finite by Exercise 1 above. Assume, as our induction hypothesis that \( \left( {\forall a}\right) \lbrack \bar{a} = n \land \operatorname{Fi... | No |
Proposition 10.32. \( \operatorname{Fin}\left( \alpha \right) \leftrightarrow \alpha \in \omega \) . | Proof. If \( \alpha \) is finite, then \( \left( {\exists n}\right) \left\lbrack {n \simeq \alpha }\right\rbrack \) . But by Proposition 10.22, this implies that \( \alpha = n \) and so \( \alpha \in \omega \) .\n\nConversely if \( \alpha \in \omega \), then since \( \alpha \simeq \alpha \) it follows that \( \alpha \)... | Yes |
Proposition 10.37. \( \bar{\alpha } < \bar{\beta } \leftrightarrow \alpha < \bar{\beta } \) . | *Proof. Since \( \bar{\alpha } \leqq \alpha \) it follows that if \( \alpha < \bar{\beta } \), then \( \bar{\alpha } < \bar{\beta } \) . Furthermore if \( \bar{\beta } \leqq \alpha \), then by *Proposition 10.13 and *Proposition 10.11,\n\n\[ \bar{\beta } = \overline{\overline{\left( \bar{\beta }\right) }} \leqq \bar{\a... | No |
Proposition 10.38. \( \alpha > 1 \rightarrow \overline{\overline{\alpha + 1}} \leqq \overline{\alpha \times \alpha } \) . | *Proof. Let \( g \) be a function defined on \( \alpha + 1 \) by\n\n\[ \n{g}^{i}\beta = \langle 0,\beta \rangle ,\;\beta < \alpha \n\]\n\n\[ \n{g}^{c}\alpha = \langle 1,0\rangle \n\]\n\nThen \( g : \left( {\alpha + 1}\right) \overset{1 - 1}{ \rightarrow }\alpha \times \alpha \) and hence \( \left( {\alpha + 1}\right) \... | Yes |
Proposition 10.43. \( \\mathcal{M}\\left( {a}^{b}\\right) \) . | Proof. \( {a}^{b} \\subseteq \\mathcal{P}\\left( {b \\times a}\\right) \) . | No |
Proposition 10.44. \( {2}^{a} \simeq \mathcal{P}\left( a\right) \) . | Proof. We define a function \( h \) on \( {2}^{a} \) in the following way\n\n\[ \n{h}^{i}f = \left\{ {x \in a \mid {f}^{i}x = 1}\right\} ,\;f \in {2}^{a}.\n\]\n\nThen \( h : {2}^{a} \rightarrow \mathcal{P}\left( a\right) \) . We wish to prove that \( h \) is one-to-one and onto.\n\nIf \( b \in \mathcal{P}\left( a\right... | Yes |
Proposition 10.45. \( {\left( {a}^{b}\right) }^{c} \simeq {a}^{b \times c} \) . | Proof. We define a function \( F \) on \( {\left( {a}^{b}\right) }^{c} \) in the following way. If \( f \in {\left( {a}^{b}\right) }^{c} \) and \( y \in c \), then \( {f}^{c}y \in {a}^{b} \), that if, \( {f}^{c}y : b \rightarrow a \) . Thus if \( x \in b \), then \( {\left( {f}^{c}y\right) }^{c}x \in a \) . We then spe... | Yes |
Proposition 11.2. \( \operatorname{cof}\left( {\alpha ,\beta }\right) \rightarrow \left\lbrack {\alpha \in {K}_{\mathrm{{II}}} \leftrightarrow \beta \in {K}_{\mathrm{{II}}}}\right\rbrack \) . | The proof is left to the reader. | No |
Proposition 11.3. \( \alpha \in {K}_{\mathrm{{II}}} \land \operatorname{cof}\left( {\alpha ,\beta }\right) \rightarrow \left( {\exists f}\right) \left\lbrack {f{\mathcal{F}}_{\mathcal{H}}\beta \land \alpha = \cup \left( {{f}^{\alpha }\beta }\right) .}\right\rbrack \) | The proof is left to the reader. | No |
Proposition 11.4. \( \operatorname{cof}\left( {\alpha ,\alpha }\right) \) | The proof is left to the reader. | No |
Proposition 11.6. \( \alpha \in {K}_{\mathrm{{II}}} \rightarrow \operatorname{cof}\left( {{\aleph }_{\alpha },\alpha }\right) \) . | The proof is left to the reader. | No |
Proposition 11.7. \[ \beta \leqq \alpha \land \left( {\exists f}\right) \lbrack f : \beta \rightarrow \alpha \land \left( {\forall \gamma < \alpha }\right) \left( {\exists \delta < \beta }\right) \left\lbrack {{f}^{i}\delta \geqq \gamma }\right\rbrack \rightarrow \left( {\exists \eta \leqq \beta }\right) \left\lbrack {... | Proof. If \( a = \left\{ {\delta < \beta \mid \left( {\forall \gamma < \delta }\right) \left\lbrack {{f}^{c}\gamma < {f}^{c}\delta }\right\rbrack }\right\} \), then \( a \subseteq \beta \) . Therefore \[ \left( {\exists \eta \leqq \beta }\right) \left( {\exists h}\right) \left\lbrack {h{\operatorname{Isom}}_{E, E}\left... | Yes |
Corollary 11.8. \( \beta \leqq \alpha \land \beta \simeq \alpha \rightarrow \left( {\exists \eta \leqq \beta }\right) \left\lbrack {\operatorname{cof}\left( {\alpha ,\eta }\right) }\right\rbrack \) . | Proof. If \( \beta \simeq \alpha \), then \( \left( {\exists f}\right) \left\lbrack {f : \beta \xrightarrow[\text{onto }]{1 - 1}\alpha }\right\rbrack \) and so \( \left( {\forall \gamma < \alpha }\right) \left( {\exists \delta < \beta }\right) \left\lbrack {{f}^{\prime }\delta = \gamma }\right\rbrack \) . Therefore, by... | Yes |
Proposition 11.9. \( \\operatorname{cof}\\left( {\\alpha ,\\beta }\\right) \\land \\operatorname{cof}\\left( {\\\alpha ,\\gamma }\\right) \\land \\gamma \\leqq \\beta \\rightarrow \\left( {\\exists \\eta \\leqq \\gamma }\\right) \\left\\lbrack {\\operatorname{cof}\\left( {\\beta ,\\eta }\\right) }\\right\\rbrack \) . | Proof. If \( \\alpha \) is confinal with \( \\beta \) and with \( \\gamma \), then\n\n\[ \n\\left( {\\exists f}\\right) \\left\\lbrack {f : \\beta \\rightarrow \\alpha \\land \\left( {\\forall \\tau < \\alpha }\\right) \\left( {\\exists \\delta < \\beta }\\right) \\left\\lbrack {{f}^{\\iota }\\delta \\geqq \\tau }\\rig... | Yes |
Proposition 11.11. \( \operatorname{cf}\left( \alpha \right) \in N \) . | Proof. Suppose that \( \beta = \operatorname{cf}\left( \alpha \right) \) . To prove that \( \beta \) is a cardinal we need only prove that if \( \gamma \simeq \beta \) then \( \gamma \geqq \beta \) . We argue by contradiction. Suppose that \( \gamma < \beta \) . Then by Corollary 11.8 \( \left( {\exists \eta \leqq \gam... | Yes |
Proposition 11.12. \( \alpha \in {N}^{\prime } \rightarrow \operatorname{cf}\left( \alpha \right) \in {N}^{\prime } \) . | Proof. If \( \alpha \in {N}^{\prime } \), then \( \alpha \in {K}_{\mathrm{{II}}} \) . Since \( \alpha \) is cofinal with \( \operatorname{cf}\left( \alpha \right) \) it follows, from Proposition 11.2 that \( \mathrm{{cf}}\left( \alpha \right) \in {K}_{\mathrm{{II}}} \) . But then \( \mathrm{{cf}}\left( \alpha \right) \... | Yes |
Proposition 11.13. \( \alpha \in {K}_{\mathrm{{II}}} \rightarrow \mathrm{{cf}}\left( \alpha \right) = \mathrm{{cf}}\left( {\aleph }_{\alpha }\right) \) . | Proof. If \( \alpha \in {K}_{\mathrm{{II}}} \), then \( {\aleph }_{\alpha } \) is cofinal with \( \alpha \) by Proposition 11.6. Since \( \alpha \) is cofinal with \( \mathrm{{cf}}\left( \alpha \right) \) it follows from Proposition 11.5 that \( {\aleph }_{\alpha } \) is cofinal with \( \operatorname{cf}\left( \alpha \... | Yes |
Proposition 11.16. \( \left\lbrack {f : a \rightarrow N}\right\rbrack \rightarrow \cup \left( {{f}^{a}a}\right) \in N \) . | Proof. Since \( \mathcal{D}\left( f\right) \) is a set, \( {f}^{\alpha }a \) is a set, indeed it is a set of ordinals. If \( \beta = \) \( \cup \left( {{f}^{\alpha }a}\right) \), then \( \bar{\beta } \leqq \beta \) . We wish to prove that \( \bar{\beta } = \beta \) . Suppose not. Suppose that \( \bar{\beta } < \beta \)... | Yes |
Corollary 11.17. \[ \left\lbrack {f : a \rightarrow N}\right\rbrack \land \left( {\exists x \in a}\right) \left\lbrack {{f}^{c}x \in {N}^{\prime }}\right\rbrack \rightarrow \cup \left( {{f}^{cc}a}\right) \in {N}^{\prime }. \] | Proof. By Proposition 11.16, \( \cup \left( {{f}^{\alpha }a}\right) \in N \) . If \( \left( {\exists x \in a}\right) \left\lbrack {{f}^{i}x \in {N}^{\prime }}\right\rbrack \), then since \( {f}^{c}x \leqq \cup \left( {{f}^{a}a}\right) \) it follows that \( \cup \left( {{f}^{a}a}\right) \in {N}^{\prime } \) . | No |
Theorem 11.20. \( \mathrm{{GCH}} \rightarrow \left\lbrack {{\operatorname{Inacc}}_{w}\left( {\aleph }_{\alpha }\right) \leftrightarrow \operatorname{Inacc}\left( {\aleph }_{\alpha }\right) }\right\rbrack \) . | Proof. By definition \( {\aleph }_{\alpha } \) inaccessible implies \( {\aleph }_{\alpha } \) weakly inaccessible. Conversely\n\n\[ \operatorname{Inf}\left( x\right) \land \overline{\bar{x}} < {\aleph }_{\alpha } \rightarrow \left( {\exists \beta }\right) \left\lbrack {\overline{\bar{x}} = {\aleph }_{\beta } \land \bet... | No |
Proposition 11.25. \( \mathcal{M}\left( {\mathop{\prod }\limits_{{x \in a}}{c}^{i}x}\right) \) . | Proof. \( \mathop{\prod }\limits_{{x \in a}}{c}^{c}x \cong {\left\lbrack \cup \left( {c}^{a}a\right) \right\rbrack }^{a} \) . | No |
Theorem 11.31. \( \mathrm{{GCH}} \rightarrow \overline{\overline{{\aleph }_{\alpha }^{{\aleph }_{\beta }}}} = {\aleph }_{\alpha }\; \) if \( {\aleph }_{\beta } < \mathrm{{cf}}\left( {\aleph }_{\alpha }\right) \)\n\n\[ = {\aleph }_{\alpha + 1}\;\text{ if }\mathrm{{cf}}\left( {\aleph }_{\alpha }\right) \leqq {\aleph }_{\... | Proof. If \( {\aleph }_{\beta } < \operatorname{cf}\left( {\aleph }_{\alpha }\right) \) then \( \alpha \neq 0 \) . If \( \left( {\exists \gamma }\right) \left\lbrack {\alpha = \gamma + 1}\right\rbrack \) then \( {\aleph }_{\alpha } \) is regular and hence\n\n\[ {\aleph }_{\beta } < \operatorname{cf}\left( {\aleph }_{\a... | Yes |
Proposition 12.5. (1) If \( \varphi \) is closed then\n\n\[ \left\lbrack {A \vDash \varphi }\right\rbrack \leftrightarrow {\varphi }^{A} \]\n\n(2) If all free variables occurring in \( \varphi \) are among \( {a}_{1},\ldots ,{a}_{n} \) then\n\n\[ {a}_{1} \in A \land \cdots \land {a}_{n} \in A \rightarrow \left\lbrack {... | Proof. We consider (1) to be the special case of (2) with \( n = 0 \), and so we need only prove (2). This we do by induction on the number of logical symbols in \( \varphi \) . We assume that \( {a}_{1},\ldots ,{a}_{n} \in A \) .\n\nIf \( \varphi \) is of the form \( a \in b \) and if \( a \) and \( b \) are among \( ... | Yes |
Proposition 13.5. If \( \varphi \) Abs \( A \), if \( {a}_{1},\ldots ,{a}_{m},{b}_{1},\ldots ,{b}_{n} \) is a list of distinct variables containing all of the free variables in \( \varphi \) and if\n\n\[ \n{b}_{1},\ldots ,{b}_{n} \in A \land \varphi \left( {{a}_{1},\ldots ,{a}_{m},{b}_{1},\ldots ,{b}_{n}}\right) \right... | Proof. Since\n\n\[ \n{b}_{1},\ldots ,{b}_{n} \in A \land \varphi \left( {{a}_{1},\ldots ,{a}_{m},{b}_{1},\ldots ,{b}_{n}}\right) \rightarrow {a}_{1},\ldots ,{a}_{m} \in A \n\]\n\nwe have that\n\n\[ \n{b}_{1},\ldots ,{b}_{n} \in A \rightarrow \left\lbrack {\left( {\exists {x}_{1},\ldots ,{x}_{m}}\right) \varphi \leftrig... | Yes |
Proposition 13.6. If \( \vdash \left\lbrack {\varphi \leftrightarrow \psi }\right\rbrack \) and if \( A \) is a nonempty class that satisfies each nonlogical axiom in some proof of \( \varphi \leftrightarrow \psi \) then\n\n\[ \varphi \text{ Abs }A \leftrightarrow \psi \text{ Abs }A\text{. } \] | Proof. If all of the free variables of \( \varphi \) and of \( \psi \) are among \( {b}_{1},\ldots ,{b}_{n} \) then by Theorem 12.6\n\n\[ {b}_{1},\ldots ,{b}_{n} \in A \rightarrow \left\lbrack {{\varphi }^{A} \leftrightarrow {\psi }^{A}}\right\rbrack . \]\n\nTherefore if \( {b}_{1},\ldots ,{b}_{n} \in A \)\n\n\[ \left\... | Yes |
Theorem 13.7. If \( \vdash \left( {\exists x}\right) \varphi \left( x\right) \), if \( A \) is a nonempty class that satisfies each nonlogical axiom in some proof of \( \left( {\exists x}\right) \varphi \left( x\right) \), and if \( \varphi \left( x\right) \) Abs \( A \) then\n\n\[ \left( {\exists x}\right) \varphi \le... | Proof. If all of the free variables of \( \left( {\exists x}\right) \varphi \left( x\right) \) are among \( {a}_{1},\ldots ,{a}_{n} \) then by Theorem 12.6\n\n\[ {a}_{1},\ldots ,{a}_{n} \in A \rightarrow \left( {\exists x \in A}\right) {\varphi }^{A}\left( x\right) . \]\n\nThen\n\n\[ {a}_{1},\ldots ,{a}_{n} \in A \righ... | Yes |
Theorem 13.8. If \( \vdash \left\lbrack {\varphi \leftrightarrow \left( {\forall x}\right) \psi \left( x\right) }\right\rbrack \) and \( \vdash \left\lbrack {\varphi \leftrightarrow \left( {\exists x}\right) \eta \left( x\right) }\right\rbrack \), if \( A \) is a nonempty class that satisfies each nonlogical axiom in s... | Proof. If all of the free variables of \( \varphi ,\left( {\forall x}\right) \psi \left( x\right) \), and \( \left( {\exists x}\right) \eta \left( x\right) \) are among \( {a}_{1},\ldots ,{a}_{n} \) then by Theorem 12.6 \[ {a}_{1},\ldots ,{a}_{n} \in A \rightarrow \left\lbrack {{\varphi }^{A} \leftrightarrow \left( {\f... | Yes |
Proposition 13.10. If \( A \neq 0 \), if \( \varphi \left( {{b}_{1},\ldots ,{b}_{n}}\right) \) Abs \( A \) and if \( \mathcal{M}\left( {B}_{1}\right) \land \cdots \) \( \land \mathcal{M}\left( {B}_{n}\right) \land {b}_{1} = {B}_{1} \) Abs \( A \land \cdots \land {b}_{n} = {B}_{n} \) Abs \( A \) then \( \varphi \left( {... | Proof (By induction on \( n \) ). If \( n = 1 \) then\n\n\[ \varphi \left( {B}_{1}\right) \leftrightarrow \left( {\forall {x}_{1}}\right) \left\lbrack {{x}_{1} = {B}_{1} \rightarrow \varphi \left( {b}_{1}\right) }\right\rbrack \]\n\n\[ \leftrightarrow \left( {\exists {x}_{1}}\right) \left\lbrack {{x}_{1} = {B}_{1} \lan... | No |
Proposition 13.14. If \( \varphi \left( x\right) \) Abs \( A \) and if all of the free variables of \( \varphi \left( x\right) \) are among \( {a}_{1},\ldots ,{a}_{n}, x \) then\n\n\[ \n{a}_{1},\ldots ,{a}_{n} \in A \rightarrow \left\lbrack {\{ x \mid \varphi \left( x\right) {\} }^{A} = \{ x \mid \varphi \left( x\right... | Proof. If \( {a}_{1},\ldots ,{a}_{n} \in A \) then since \( \varphi \left( x\right) \) is absolute with respect to \( A \)\n\n\[ \nx \in \{ x \mid \varphi \left( x\right) {\} }^{A} \leftrightarrow x \in A \land {\varphi }^{A}\left( x\right)\n\]\n\n\[ \n\leftrightarrow x \in A \land \varphi \left( x\right)\n\]\n\n\[ \n\... | Yes |
Proposition 13.15. If \( A \) is nonempty and transitive and if \( \{ y \mid \varphi \left( y\right) \} \) is a set, then\n\n\[ \left\lbrack {a = \{ y \mid \varphi \left( y\right) \} }\right\rbrack \text{ Abs }A\;\text{ iff }\{ y \mid \varphi \left( y\right) \} \text{ Abs }A. \] | Proof. If all of the free variables of \( \varphi \left( y\right) \) are among \( {a}_{1},\ldots ,{a}_{n}, y \) then\n\n\[ \left\lbrack {a = \{ y \mid \varphi \left( y\right) \} }\right\rbrack \text{ Abs }A \]\n\n\[ \leftrightarrow {a}_{1},\ldots ,{a}_{n}, a \in A \rightarrow \left\lbrack {\left( {\forall y}\right) \le... | Yes |
Proposition 13.16. If \( \varphi \) is quantifier free then \( \varphi \) Abs \( A \) . | Proof. If \( \varphi \) is quantifier free then \( {\varphi }^{A} \leftrightarrow \varphi \) . | No |
Proposition 13.17. \( a \in b \) Abs \( A \) . | Proof. The formula \( a \in b \) is quantifier free. | No |
Proposition 13.18. If \( A \) is nonempty and transitive then\n\n(1) \( a \subseteq b \) Abs \( A \) ,\n\n(2) \( a = b \) Abs \( A \) . | Proof.\n\n(1) \( a \subseteq b \leftrightarrow \left( {\forall x}\right) \left\lbrack {x \in a \rightarrow x \in b}\right\rbrack \) .\n\nSince \( A \) is transitive \( a \in A \) implies \( a \subseteq A \), i.e., \( x \in a \land a \in A \) implies \( x \in A \) . From Propositions 13.17, 13.3, and 13.6 it then follow... | Yes |
Proposition 13.19. If \( A \) is nonempty and transitive then\n\n(1) \( 0\mathrm{{Abs}}A \) ,\n\n(2) \( \left\lbrack {a \cup b}\right\rbrack \) Abs \( A \) ,\n\n(3) \( \{ a, b\} \) Abs \( A \) ,\n\n(4) \( \cup \left( a\right) \) Abs \( A \) ,\n\n(5) \( \left\lbrack {a - b}\right\rbrack \) Abs \( A \) . | Proof. (1) Since \( a \neq a \) Abs \( A \) we have\n\n\[{0}^{A} = \left\{ {x \in A \mid {\left\lbrack x \neq x\right\rbrack }^{A}}\right\} = \{ x \in A \mid x \neq x\} = 0.\]\n\n(2) If \( a, b \in A \) then\n\n\[{\left\lbrack a \cup b\right\rbrack }^{A} = \left\{ {x \in A \mid {\left\lbrack x \in a \vee x \in b\right\... | Yes |
Proposition 13.20. If \( A \) is nonempty and transitive then\n\n(1) \( \operatorname{Tr}\left( a\right) \) Abs \( A \) ,\n\n(2) Ord (a) Abs \( A \) . | Proof.\n\n(1) \( \operatorname{Tr}\left( a\right) \leftrightarrow \left( {\forall x}\right) \left\lbrack {x \in a \rightarrow x \subset a}\right\rbrack \) .\n\n(2) \( \operatorname{Ord}\left( a\right) \leftrightarrow \operatorname{Tr}\left( a\right) \land \n\n\\left( {\forall x, y}\right) \left\lbrack {x \in a \land y ... | No |
Proposition 13.21. If \( A \) is nonempty and transitive then\n\n(1) \( \left\lbrack {a \in \omega }\right\rbrack \) Abs \( A \) ,\n\n(2) \( \left\lbrack {a = \omega }\right\rbrack \) Abs \( A \) if \( \omega \subseteq A \) . | Proof.\n\n(1) \( a \in \omega \leftrightarrow a \cup \{ a\} \subseteq {K}_{1} \)\n\n\[ \leftrightarrow \left( {\forall x}\right) \left\lbrack {x \in a \vee x = a \rightarrow x \in {K}_{1}}\right\rbrack \]\n\n\[ \leftrightarrow \left( {\forall x}\right) \lbrack x \in a \vee x = a \rightarrow x = 0 \vee \]\n\n\[ \left( {... | Yes |
Proposition 13.22. If \( A \) is nonempty and transitive then\n\n(1) \( \left\lbrack {\alpha < \beta }\right\rbrack \) Abs \( A \) ,\n\n(2) \( \left\lbrack {\alpha = \beta }\right\rbrack \) Abs \( A \) ,\n\n(3) \( \left\lbrack {\gamma = \max \left( {\alpha ,\beta }\right) }\right\rbrack \) Abs \( A \) . | Proof.\n\n(1) \( \left\lbrack {\alpha < \beta }\right\rbrack \leftrightarrow \operatorname{Ord}\left( \alpha \right) \land \operatorname{Ord}\left( \beta \right) \land \alpha \in \beta \) .\n\n(2) \( \left\lbrack {\alpha = \beta }\right\rbrack \leftrightarrow \operatorname{Ord}\left( \alpha \right) \land \operatorname{... | Yes |
Proposition 13.23. If \( A \) is nonempty and transitive then\n\n(1) \( \left\lbrack {\langle \alpha ,\beta \rangle \operatorname{Le}\langle \gamma ,\delta \rangle }\right\rbrack \) Abs \( A \) ,\n\n(2) \( \left\lbrack {\langle \alpha ,\beta \rangle {R}_{0}\langle \gamma ,\delta \rangle }\right\rbrack \) Abs \( A \) . | (1) \( \left\lbrack {\langle \alpha ,\beta \rangle \operatorname{Le}\langle \gamma ,\delta \rangle }\right\rbrack \leftrightarrow \left\lbrack {\alpha < \gamma \vee \left\lbrack {\alpha = \gamma \land \beta < \delta }\right\rbrack }\right\rbrack \) .\n\n(2) \( \left\lbrack {\langle \alpha ,\beta \rangle {R}_{0}\langle ... | Yes |
Proposition 13.24. \( A \neq 0 \land \operatorname{Tr}\left( A\right) \rightarrow \operatorname{STM}\left( {A,\text{ Ax. }1}\right) \) . | Proof. Since Axiom 1 assures us that\n\n(1) \( x = y \land x \in z \rightarrow y \in z \)\n\nholds for all \( x, y \), and \( z \), it also assures us that (1) holds for all \( x, y, z \in A \), that is\n\n\[ \left( {\forall x, y, z \in A}\right) \left\lbrack {x = y \land x \in z \rightarrow y \in z}\right\rbrack .\n\]... | Yes |
Proposition 13.25. If \( A \) is nonempty and transitive then \( A \) is a standard transitive model of the Axiom of Pairing iff\n\n\[ \left( {\forall x, y \in A}\right) \left\lbrack {\{ x, y\} \in A}\right\rbrack \text{.} \] | Proof. Ax. \( 2 \leftrightarrow \left( {\forall x, y}\right) \left( {\exists z}\right) \left\lbrack {z = \{ x, y\} }\right\rbrack \) . Since \( A \) is nonempty and transitive \( A \) is a standard transitive model of Axiom 2 if and only if\n\n(1) \( \left( {\forall x, y \in A}\right) \left( {\exists z \in A}\right) {\... | Yes |
Proposition 13.26. \( \operatorname{STM}\left( {A,\mathrm{{Ax}}.2}\right) \rightarrow \langle a, b\rangle \) Abs \( A \) . | Proof. If \( A \) is a standard transitive model of Axiom 2 and if \( a, b \in A \), then \( \{ a\} \in A \) and \( \{ a, b\} \in A \) . Then\n\n\[ \n{\left\lbrack \langle a, b\rangle \right\rbrack }^{A} = \left\{ {x \in A \mid {\left\lbrack x = \{ a\} \vee x = \{ a, b\} \right\rbrack }^{A}}\right\} \n\]\n\n\[ \n= \{ x... | Yes |
Proposition 13.27. If \( A \) is a nonempty, transitive model of the Axiom of Pairing then\n\n(1) \( \operatorname{Rel}\left( a\right) \) Abs \( A \) ,\n\n(2) \( {\mathcal{U}}_{n}\left( a\right) \) Abs \( A \) ,\n\n(3) \( {u}_{{n}_{2}}\left( a\right) \) Abs \( A \) ,\n\n(4) \( {\mathcal{F}}_{nc}\left( a\right) \) Abs \... | Proof.\n\n(1) \( \mathcal{R}{el}\left( a\right) \leftrightarrow \left( {\forall x}\right) \left\lbrack {x \in a \rightarrow \left( {\exists y, z}\right) \left\lbrack {x = \langle y, z\rangle }\right\rbrack }\right\rbrack \) .\n\n(2) \( {\mathcal{U}}_{n}\left( a\right) \leftrightarrow \left( {\forall x, y, z}\right) \le... | Yes |
Theorem 13.28. If \( A \) is a nonempty transitive model of the \( A \) xiom of Pairing then\n\n(1) \( a \times b \) Abs \( A \) ,\n\n(2) \( {a}^{-1}\mathrm{{Abs}}A \) ,\n\n(3) \( \mathcal{D}\left( a\right) \) Abs \( A \) ,\n\n(4) \( \mathcal{W}\left( a\right) \) Abs \( A \) ,\n\n(5) \( {a}^{c}b \) Abs \( A \) ,\n\n(6)... | Proof. If \( a, b \in A \) then\n\n(1) \( {\left\lbrack a \times b\right\rbrack }^{A} = \left\{ {x \in A \mid \left( {\exists y, z \in A}\right) {\left\lbrack y \in a \land z \in b \land x = \langle y, z\rangle \right\rbrack }^{A}}\right\} \n\n\[ = \{ x \in A \mid \left( {\exists y, z \in A}\right) \left\lbrack {y \in ... | No |
Proposition 13.29. If \( A \) is a nonempty transitive model of the Axiom of Pairing then\n\n(1) \( f{\mathcal{F}}_{\mathcal{n}}a \) Abs \( A \) ,\n\n(2) \( f{\mathcal{F}}_{{n}_{2}}a\mathrm{{Abs}}A \) . | Proof\n\n(1) \( f{\mathcal{F}}_{n}a \leftrightarrow {\mathcal{F}}_{n}c\left( f\right) \land \mathcal{D}\left( f\right) = a \) .\n\n(2) \( f{\mathcal{F}}_{{n}_{2}}a \leftrightarrow {\mathcal{F}}_{n}{c}_{2}\left( f\right) \land \mathcal{D}\left( f\right) = a \) . | Yes |
Proposition 13.30. If \( A \) is a nonempty transitive model of the Axiom of Pairing and if \( a{R}_{1}b \) and \( a{R}_{2}b \) are each absolute with respect to \( A \) then \( f{\operatorname{Isom}}_{{R}_{1},{R}_{2}}\left( {a, b}\right) \) Abs \( A \) . | \[ f{\operatorname{Isom}}_{{R}_{1},{R}_{2}}\left( {a, b}\right) \leftrightarrow f{\mathcal{F}}_{{n}_{2}}a \land \mathcal{W}\left( f\right) = b \] \[ \land \left( {\forall x, y}\right) \left\lbrack {x \in a \land y \in a\land \langle x, y\rangle \in {R}_{1} \rightarrow \left\langle {{f}^{i}x,{f}^{i}y}\right\rangle \in {... | Yes |
Theorem 13.31. If \( A \) is nonempty and transitive then \( A \) is a standard transitive model of the Axiom of Unions iff\n\n\[ \left( {\forall x \in A}\right) \left\lbrack {\cup \left( x\right) \in A}\right\rbrack \text{.} \] | Proof. Ax. \( 3 \leftrightarrow \left( {\forall x}\right) \left( {\exists y}\right) \left\lbrack {y = \cup \left( x\right) }\right\rbrack \) .\n\nSince \( A \) is nonempty and transitive we have\n\n\[ \operatorname{STM}\left( {A,\text{ Ax. }3}\right) \]\n\nif and only if\n\n\[ \left( {\forall x \in A}\right) \left( {\e... | Yes |
Proposition 13.32. If \( A \) is nonempty and transitive then \( A \) is a standard transitive model of the Axiom of Powers iff\n\n\[ \left( {\forall x \in A}\right) \left\lbrack {\mathcal{P}\left( x\right) \cap A \in A}\right\rbrack . \] | Proof. Ax. \( 4 \leftrightarrow \left( {\forall x}\right) \left( {\exists y}\right) \left( {\forall z}\right) \left\lbrack {z \in y \leftrightarrow z \subseteq x}\right\rbrack \) . Since \( A \) is nonempty and transitive we have\n\n\[ \operatorname{STM}\left( {A,\text{ Ax. }4}\right) \]\n\nif and only if\n\n\[ \left( ... | Yes |
Proposition 13.33. If \( A \) is nonempty and transitive then \( A \) is a standard transitive model of Axiom \( {5}_{\varphi } \), iff for \( a \in A \)\n\n\[ \n\\left( {\\forall x, y, z \\in A}\\right) \\left\\lbrack {{\\varphi }^{A}\\left( {x, y}\\right) \\land {\\varphi }^{A}\\left( {x, z}\\right) \\rightarrow y = ... | Proof. Since \( A \) is nonempty and transitive \( A \) is a standard transitive model of Axiom \( {5}_{\varphi } \) iff for \( a \\in A \)\n\n\[ \n\\left( {\\forall x, y, z \\in A}\\right) \\left\\lbrack {{\\varphi }^{A}\\left( {x, y}\\right) \\land {\\varphi }^{A}\\left( {x, z}\\right) \\rightarrow y = z}\\right\\rbr... | Yes |
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