Question

Argue that \(E=E F \cup E F^{c}, E \cup F=E \cup F E^{c}\).

Short answer

We can prove that \(E = EF \cup EF^c\) and \(E \cup F = E \cup FE^c\) using set identities and logical equivalence. To prove \(E = EF \cup EF^c\), we need to show that \(E \subseteq EF \cup EF^c\) and \(EF \cup EF^c \subseteq E\). For any element \(x \in E\), it can either belong to \(F\) or \(F^c\), resulting in \(x \in EF \cup EF^c\); hence, \(E \subseteq EF \cup EF^c\). Conversely, since elements of \(EF\) and \(EF^c\) belong to \(E\), we have \(EF \cup EF^c \subseteq E\). Thus, \(E = EF \cup EF^c\). Now, to prove \(E \cup F = E \cup FE^c\), we need to show that \(E \cup F \subseteq E \cup FE^c\) and \(E \cup FE^c \subseteq E \cup F\). If \(x \in E \cup F\), then \(x \in E\) or \(x \in F\), and we can conclude that \(x \in E \cup FE^c\); hence, \(E \cup F \subseteq E \cup FE^c\). Similarly, if \(x \in E \cup FE^c\), then \(x \in E\) or \(x \in FE^c\), resulting in \(x \in E \cup F\); thus, \(E \cup FE^c \subseteq E \cup F\). Consequently, \(E \cup F = E \cup FE^c\).

Step-by-step solution

Prove that \( E \subseteq EF \cup EF^c\)

To demonstrate that \(E \subseteq EF \cup EF^c\), we have to show that every element of \(E\) also belongs to \(EF \cup EF^c\). For any element \(x\) in \(E\), since \(x \in E\), it can either belong to \(F\) or \(F^c\). If \(x \in F\), then \(x \in EF\), and if \(x \in F^c\), then \(x \in EF^c\). In both cases, \(x \in EF \cup EF^c\), which means that \( E \subseteq EF \cup EF^c\).

Prove that \(EF \cup EF^c \subseteq E\)

To demonstrate that \(EF \cup EF^c \subseteq E\), we have to show that every element, \(x\), in the union also belongs to \(E\). If \(x \in EF\), then clearly \(x \in E\), since \(EF\) is the intersection of the sets \(E\) and \(F\). Similarly, if \(x \in EF^c\), then \(x\) must belong to \(E\), since \(EF^c\) is the intersection of the sets \(E\) and \(F^c\). Therefore, \(EF \cup EF^c \subseteq E\). Since we have proven both that \( E \subseteq EF \cup EF^c\) and \(EF \cup EF^c \subseteq E\), we can conclude that \(E = EF \cup EF^c\). Now let's prove the second statement: \(E \cup F = E \cup FE^c\).

Prove that \(E \cup F \subseteq E \cup FE^c\)

To show that \(E \cup F \subseteq E \cup FE^c\), we need to demonstrate that every element of the left-hand side belongs to the right-hand side as well. Let \(x\) be an element in \(E \cup F\), meaning \(x \in E\) or \(x \in F\). If \(x \in E\), then of course \(x \in E \cup FE^c\). If \(x \in F\), then \(x\) can either belong to \(E\) or \(E^c\). If \(x \in E\), then \(x \in E \cup FE^c\), and if \(x \in E^c\), then \(x \in FE^c \subseteq E \cup FE^c\). Therefore, \(E \cup F \subseteq E \cup FE^c\).

Prove that \(E \cup FE^c \subseteq E \cup F\)

To show that \(E \cup FE^c \subseteq E \cup F\), we need to demonstrate that every element of the left-hand side also belongs to the right-hand side. Let \(x\) be an element in \(E \cup FE^c\), meaning \(x \in E\) or \(x \in FE^c\). If \(x \in E\), then of course \(x \in E \cup F\). If \(x \in FE^c\), then \(x\) is in both \(F\) and \(E^c\), meaning \(x \in F\), and therefore \(x \in E \cup F\). Hence, \(E \cup FE^c \subseteq E \cup F\). Since we have proven that \( E \cup F \subseteq E \cup FE^c\) and \(E \cup FE^c \subseteq E \cup F\), we can conclude that \(E \cup F = E \cup FE^c\).