Show that the definition of equality in Definition 3.1.4 is reflexive, symmetric, and transitive.
- \(\bullet\) Show that Definition 3.1.4 is reflexive, symmetric, and transitive.
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\(\Vdash\) { prove by cases }
- \(\bullet\) Show that Definition 3.1.4 is reflexive, i.e., \(A=A\)
- \(\Vdash\) \(A=A\)
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\(\equiv\) { \(x\in A \iff x\in A\) }
- \(\text{True}\)
- \(\square\)
- \(\bullet\) Show that Definition 3.1.4 is symmetric, i.e., \(A=B\implies B=A\)
- \(\Vdash\) \(A=B\)
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\(\equiv\) { Definition 3.1.4 }
- \((x\in A\implies x\in B)\land (x\in B\implies x\in A)\)
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\(\equiv\) { \(P\land Q\equiv Q\land P\) }
- \((x\in B\implies x\in A)\land (x\in A\implies x\in B)\)
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\(\equiv\) { Definition 3.1.4 }
- \(B=A\)
- \(\square\)
- \(\bullet\) Show that Definition 3.1.4 is transitive, i.e., \((A=B)\land(B=C)\implies A=C\)
- \(\Vdash\) \((A=B)\land(B=C)\)
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\(\equiv\) { Definition 3.1.4 }
- \((x\in A\implies x\in B)\land (x\in B\implies x\in A)\land (x\in B\implies x\in C)\land (x\in C\implies x\in B)\)
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\(\equiv\) { \(P\land Q\equiv Q\land P\) }
- \((x\in A\implies x\in B)\land (x\in B\implies x\in C)\land (x\in C\implies x\in B)\land (x\in B\implies x\in A)\)
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\(\Rightarrow\) { \([(P\implies Q)\land (Q\implies R)]\implies (P\implies R)\) }
- \((x\in A \implies x\in C)\land (x\in C\implies x\in A)\)
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\(\equiv\) { Definition 3.1.4 }
- \(A=C\)
- \(\square\)
- \(\square\)