Exercise 3.1.2 Using only Definition 3.1.4, Axiom 3.1, Axiom 3.2, and Axiom 3.3, prove that the sets \(\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\), and \(\{\emptyset, \{\emptyset\}\}\) are all distinct (i.e, no two of them are equal to each other).
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\(\bullet\) Show that sets are all distinct from each other
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\(\Vdash\) { Prove by cases }
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\(\bullet\) Show that \(\emptyset \neq \{\emptyset\}\)
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\(\Vdash\) { Proof by contradiction }
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\(\bullet\) Show that there is a contradiction, when
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– \(\emptyset = \{\emptyset\}\)
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\(\Vdash \emptyset = \{\emptyset\}\)
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\(\equiv\) { Definition 3.1.4 }
- \((\forall x, x \in \emptyset \implies x \in \{\emptyset\}) \land (\forall x, x\in \{\emptyset\} \implies x \in \emptyset)\)
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\(\vdash\) { And-Elimination: \((P \land Q) \vdash Q\) }
- \((\forall x, x \in \{\emptyset\} \implies x \in \emptyset)\)
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\(\vdash\) { Substitute \(\emptyset\) into \(x\) }
- \(\emptyset \in \{\emptyset\} \implies \emptyset \in \emptyset\)
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\(\equiv\) { Axiom 3.2: \(\emptyset \notin \emptyset\) and Modus tollens: \((P \implies Q, \lnot Q) \vdash \lnot P\) }
- \(\emptyset \notin \{ \emptyset \}\)
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\(\vdash\) { \(\emptyset \in \{\emptyset\}\) by definition of \(\in\) and And-introduction \((P , Q ) \vdash P \land Q\) }
- \(\emptyset \in \{\emptyset\} \land \emptyset \notin \{\emptyset\}\)
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\(\vdash\) { \(P , \lnot P \vdash \bot\) }
- \(\bot\)
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\(\square\)
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\(\square\)
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\(\bullet\) Show that \(\emptyset \neq \{\{\emptyset\}\}\)
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\(\cdots\)
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\(\square\)
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\(\bullet\) Show that \(\emptyset \neq \{\emptyset, \{\emptyset\}\}\)
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\(\cdots\)
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\(\square\)
- \(\bullet\) Show that \(\{\emptyset\} \neq \{\{\emptyset\}\}\)
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\(\cdots\)
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\(\square\)
- \(\bullet\) Show that \(\{\emptyset\} \neq \{\emptyset, \{\emptyset\}\}\)
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\(\cdots\)
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\(\square\)
- \(\bullet\) Show that \(\{\{\emptyset\}\} \neq \{\emptyset, \{\emptyset\}\}\)
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\(\cdots\)
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\(\square\)
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\(\square\)