Use the axiom of regularity (and the singleton set axiom) to show that if \(A\) is a set, then \(A \in A\). Furthermore, show that if \(A\) and \(B\) are two sets, then either \(A \notin B\) or \(B \notin A\) (or both).
- \(\bullet\) Show that \(A \notin A\), when
- - \(A\) is a set
- - axiom 3.9: the axiom of regularity
- - axiom 3.3: the singleton set axiom
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\(\Vdash\) { If \(A\) is an set then \(A\) is an object and by axiom 3.3 }
- \(\{A\}\) exists and is a set
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\(\vdash\) { \(\{A\} \neq \emptyset\) and axiom 3.9 }
- \(\exists y (y \in \{A\} \land y \cap \{A\} = \emptyset)\)
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\(\vdash\) { \(y = A\) by axiom 3.3 and And-elimination }
- \(A \cap \{A\} = \emptyset\)
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\(\vdash\) { \(A \in A \implies A \cap \{A\} \neq \emptyset\) and moduls tollens }
- \(A \notin A\)
- \(\square\)
- \(\bullet\) Show \(A \notin B \lor B \notin A\)
- - \(A\) and \(B\) are sets
- - axiom 3.9: the axiom of regularity
- - axiom 3.3: the singleton set axiom
- - \(S \notin S\) for any set \(S\)
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\(\Vdash\) { axiom 3.3 }
- \(\{A, B\}\) exists and is a non-empty set
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\(\vdash\) { axiom 3.9 and modus ponens }
- \(\exists y ( y \in \{A, B\} \land (y \cap \{A, B\} = \emptyset))\)
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\(\vdash\) { axiom 3.3: \(\exists y ( y \in \{A, B\}) \vdash y = A \lor y = B\) }
- \((A \cap \{A, B\} = \emptyset) \lor (B \cap \{A, B\} = \emptyset)\)
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\(\vdash\) { Definition of \(\cap\) }
- \(((A \notin A) \land (B \notin A)) \lor ((A \notin B) \land (B \notin B))\)
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\(\vdash\) { \(S \notin S\) for any set \(S\) and And-elimination }
- \((B \notin A) \lor (A \notin B)\)
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\(\vdash\) { \(\lor\)-commutation }
- \((A \notin B) \lor (B \notin A)\)
- \(\square\)