91Ó°ÊÓ

In order to prove the first Absorption Law, that \(M_{1} \cap\left(M_{1} \cup M_{2}\right) = M_{1}\), we need to show that every element in \(M_{1} \cap\left(M_{1} \cup M_{2}\right)\) is in \(M_{1}\), and every element in \(M_{1}\) is in \(M_{1} \cap\left(M_{1} \cup M_{2}\right)\). Let's analyze both cases: 1. If \(x \in M_{1} \cap\left(M_{1} \cup M_{2}\right)\), then \(x \in M_{1}\) and \(x \in\left(M_{1} \cup M_{2}\right)\). Since \(x \in M_{1}\), we can say that \(x \in M_{1}\). 2. If \(x \in M_{1}\), then \(x \in M_{1}\) or \(x \in M_{2}\), which means \(x \in\left(M_{1} \cup M_{2}\right)\). Since \(x \in M_{1}\) and \(x \in\left(M_{1} \cup M_{2}\right)\), we have \(x \in M_{1} \cap\left(M_{1} \cup M_{2}\right)\). Since all elements in \(M_{1} \cap\left(M_{1} \cup M_{2}\right)\) are in \(M_{1}\), and all elements in \(M_{1}\) are in \(M_{1} \cap\left(M_{1} \cup M_{2}\right)\), we have proven the first Absorption Law.

In order to prove the second Absorption Law, that \(M_{1} \cup\left(M_{1} \cap M_{2}\right) = M_{1}\), we need to show that every element in \(M_{1} \cup\left(M_{1} \cap M_{2}\right)\) is in \(M_{1}\), and every element in \(M_{1}\) is in \(M_{1} \cup\left(M_{1} \cap M_{2}\right)\). Let's analyze both cases: 1. If \(x \in M_{1} \cup\left(M_{1} \cap M_{2}\right)\), then \(x \in M_{1}\) or \(x \in\left(M_{1} \cap M_{2}\right)\). If \(x \in\left(M_{1} \cap M_{2}\right)\), then \(x \in M_{1}\). In both cases, we can say that \(x \in M_{1}\). 2. If \(x \in M_{1}\), then \(x \in M_{1}\) or \(x \in\left(M_{1} \cap M_{2}\right)\), which means \(x \in M_{1} \cup\left(M_{1} \cap M_{2}\right)\). Since all elements in \(M_{1} \cup\left(M_{1} \cap M_{2}\right)\) are in \(M_{1}\), and all elements in \(M_{1}\) are in \(M_{1} \cup\left(M_{1} \cap M_{2}\right)\), we have proven the second Absorption Law.