Equivalence of Formulations of Axiom of Empty Set
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Theorem
The following formulations of the Axiom of the Empty Set are equivalent:
Formulation 1
- $\exists x: \forall y \in x: y \ne y$
Formulation 2
- $\exists x: \forall y: \paren {\neg \paren {y \in x} }$
Proof
Formulation 1 implies Formulation 2
\(\text {(1)}: \quad\) | \(\ds \) | \(\) | \(\ds \forall y: y = y\) | Equality is Reflexive | ||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(\) | \(\ds \neg \exists y: y \ne y\) | From $(1)$: Assertion of Universality | ||||||||||
\(\text {(3)}: \quad\) | \(\ds \) | \(\) | \(\ds x := \set { {y: y \ne y} }\) | |||||||||||
\(\text {(4)}: \quad\) | \(\ds \) | \(\) | \(\ds \neg \exists y: y \in x\) | From $(2)$ and $(3)$ |
$\Box$
Formulation 2 implies Formulation 1
\(\text {(1)}: \quad\) | \(\ds \) | \(\) | \(\ds \forall y: \paren {\neg \paren {y \in x} }\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(\) | \(\ds \forall y: \neg \paren {y \in x} \land \paren {y \in x} \implies y \ne y\) | Rule of Explosion | ||||||||||
\(\text {(3)}: \quad\) | \(\ds \) | \(\) | \(\ds \forall y: \paren {\neg \paren {y \in x} \implies \paren {y \in x \implies y \ne y} }\) | Rule of Exportation, from $(2)$ | ||||||||||
\(\text {(4)}: \quad\) | \(\ds \) | \(\) | \(\ds \forall y: \paren {y \in x \implies y \ne y}\) | Modus Ponendo Ponens, from $(1)$ and $(3)$ |
$\blacksquare$
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 0.2$. Sets