I haven't been able to find an answer to this question on here or anywhere that is kind of explicit about it. If we have a set (R - S) ∩ T, and we wanted to write it using logical operators, could we do the following and still have each step be a valid or well-defined expression? Let x be an arbitrary element in U, for which R,S and T are subsets of.
- (x∈R ^ x∉S) ∩ T
- (x∈R ^ x∉S) ^ x∈T
My confusion arises from step 1. I want to be systematic and go step by step when translating the set notation into logical operators, but I have a feeling step 1) is not a legal expression or is not allowed, as it is mixing set symbols (the set intersection symbol) with logical operators (like ^). Am I correct? Is it better to just go straight to step 2 in one step?
I appreciate any help very much!
You have the right idea, but technically you are not writing well-formed formulas. This becomes obvious when you consider the interpretation of a string such as $(x \in R \wedge x \notin S) \cap T$. The interpretation of the substring within the parentheses is a truth value, such as "True" or "False." What does it mean to take the intersection of "True" and the set $T$? It's nonsense because set theory defines intersection on sets and not truth values. As far as I know, there is no system that integrates the languages of set theory and propositional logic as you do in your example.
You can go back and forth between the languages of set theory and propositional logic using set builder notation as follows:
$ \begin{array}{llll} (R - S) \cap T \\ = (\{ x:x \in R \} - \{ x:x \in S \}) \cap \{ x:x \in T \}& \text{by definition of sets $R,S,T$} \\ = (\{ x:x \in R \} \cap \{ x:x \in S \}^C) \cap \{ x:x \in T \}& \text{by definition of set difference} \\ = (\{ x:x \in R \} \cap \{ x:x \notin S \}) \cap \{ x:x \in T \}& \text{by definition of complement} \\ = \{ x:x \in R \wedge x \notin S\} \cap \{ x:x \in T \}& \text{by definition of intersection} \\ = \{ x:x \in R \wedge x \notin S \wedge x \in T\} & \text{by definition of intersection} \\ = \{ x:x \in R \wedge \neg(x \in S) \wedge x \in T\} & \text{by definition of $\notin$} \\ \end{array} $