- By hypothesis, $S$ is a set with some subsets $A,B\subseteq S$ and the corresponding indicator functions $\chi_A$ and $\chi_B.$

- By definition, the carrier of $\max(\chi_A,\chi_B)$ is the set $\{x\mid x\in S,\max(\chi_A,\chi_B)=1\}.$
- This set equals obviously the set $\{x\mid x\in S,\chi_A(x)=1\vee \chi_B(x)=1\}.$
- But it carrier is the set union $A\cup B.$
- Therefore, $\chi_{A\cup B}=\max(\chi_A,\chi_B).$

- By definition, the carrier of $\min(\chi_A,\chi_B)$ is the set $\{x\mid x\in S,\min(\chi_A,\chi_B)=1\}.$
- This set equals obviously the set $\{x\mid x\in S,\chi_A(x)=1\wedge \chi_B(x)=1\}.$
- But it carrier is the set intersection $A\cap B.$
- Therefore, $\chi_{A\cap B}=\min(\chi_A,\chi_B).$

- The reader is invited to verify the other identities for set complement and set difference as an exercise.

q.e.d

| | | | created: 2019-09-03 22:49:28 | modified: 2019-09-03 22:49:28 | by: *bookofproofs*

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