Exercise 3.1.1

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