# ZFC Axioms of Set Theory

## Main Act 3

• Extension $$\forall x\ \forall y\ \forall z\ (z \in x \leftrightarrow z \in y)\rightarrow x=y$$
• Foundation $$\forall x[\exists y(y\in x)\rightarrow \exists y(y\in x \wedge \neg\exists z(z\in x \wedge z\in y))]$$
• Pairing $$\forall a\forall b\exists x[a\in x \wedge b\in x]$$
• Union $$\forall X\exists U[\forall Y\forall x(x\in Y \wedge Y \in X)\rightarrow x\in U]$$
• Power $$\forall X\exists P\forall z[z\subset X\rightarrow z\in P]$$
• Infinity $$\exists x[\forall z(z=\emptyset)\rightarrow z\in x \wedge \forall x\in x\forall z(z=S(x)\rightarrow z\in x)]$$
• Separation $$\forall x\forall p\exists y[\forall u(u\in y\leftrightarrow(u\in x\wedge \phi(u,p)))]$$
• Replacement $$\forall A\forall p[\forall x\in A\exists !y\phi(x, y, p)\rightarrow\exists Y\forall x\in A\exists y\in Y\phi(x, y,p)]$$
• Choice $$\forall X[\forall x\in X(x\not=\emptyset) \wedge \forall x\in X\forall y\in X(x=y\vee x\cap y=\emptyset)]\rightarrow\exists S\forall x\in X\exists !z(z\in S\wedge z\in x)$$

## Warm Up

• $$x\subset X\leftrightarrow \forall z(z\in x\rightarrow z\in X)$$
• $$S(x) = x\cup \{x\}$$
• $$\emptyset = \forall x(x!=x)$$
• $$\exists !x\phi(x)\leftrightarrow \exists x\phi(x)\wedge \forall x\forall y(\phi(x)\wedge \phi(y)\rightarrow x=y)$$

## Main Act

• $$\forall x\forall y[\forall z(z \in x \leftrightarrow z \in y)\rightarrow x=y]$$
• $$\forall x[\exists y(y\in x)\rightarrow \exists y(y\in x \wedge \neg\exists z(z\in x \wedge z\in y))]$$
• $$\forall a\forall b\exists x[a\in x \wedge b\in x]$$
• $$\forall X\exists U[\forall Y\forall x(x\in Y \wedge Y \in X)\rightarrow x\in U]$$
• $$\forall X\exists P\forall z[z\subset X\rightarrow z\in P]$$
• $$\exists x[\forall z(z=\emptyset)\rightarrow z\in x \wedge \forall x\in x\forall z(z=S(x)\rightarrow z\in x)]$$
• $$\forall x\forall p\exists y[\forall u(u\in y\leftrightarrow(u\in x\wedge \phi(u,p)))]$$
• $$\forall A\forall p[\forall x\in A\exists !y\phi(x, y, p)\rightarrow\exists Y\forall x\in A\exists y\in Y\phi(x, y,p)]$$
• $$\forall X[\forall x\in X(x\not=\emptyset) \wedge \forall x\in X\forall y\in X(x=y\vee x\cap y=\emptyset)]\rightarrow\exists S\forall x\in X\exists !z(z\in S\wedge z\in x)$$

posted 2022-03-11 | tags: set theory