conjecture 17 if A, B, and C are sets, then (A\B)U C = A\(BUC)
To prove or disprove the given conjecture, that is, if \(A\), \(B\), and \(C\) are sets, then \((A\setminus B)\cup C = A\setminus (B\cup C)\), we need to demonstrate whether the statement holds true for all possible sets \(A\), \(B\), and \(C\) or provide a counterexample to disprove it.
To prove the conjecture, we start by analyzing the left-hand side (LHS) of the equation \((A\setminus B)\cup C\). The expression \(A\setminus B\) represents the set of elements in \(A\) but not in \(B\), while the union symbol \(\cup\) combines the elements from that set with the elements in set \(C\).
Next, we examine the right-hand side (RHS) of the equation, which is \(A\setminus (B\cup C)\). Here, the expression \(B\cup C\) represents the set that includes all the elements in \(B\) and \(C\) combined. Then, \(A\setminus (B\cup C)\) denotes the set of elements in \(A\) but not in \(B\cup C\).
To show that the LHS and RHS are equivalent, we need to establish that they contain the same elements.
Suppose an element \(x\) belongs to the LHS of the equation \((A\setminus B)\cup C\). This implies that \(x\) is either an element of \(A\) but not of \(B\) or an element of \(C\). In other words, either \(x\in A\) and \(x\notin B\) or \(x\in C\).
If \(x\in A\) and \(x\notin B\), then \(x\) must also not be in \(B\cup C\). Thus, \(x\in A\setminus (B\cup C)\) as well.
On the other hand, if \(x\in C\), then \(x\) will also be present in both the LHS and RHS, as \(x\) is an element of \(C\) and hence included in the union operation.
Conversely, if an element \(x\) belongs to the RHS of the equation \(A\setminus (B\cup C)\), this indicates that \(x\) is an element of \(A\) but not present in \(B\cup C\). This implies that \(x\) is either not in \(B\) or not in \(C\), or both. Consequently, either \(x\in A\) and \(x\notin B\) or \(x\in A\) and \(x\notin C\), which satisfies the condition for the LHS as well.
Since we have demonstrated that an element \(x\) belongs to the LHS if and only if it belongs to the RHS, we can conclude that \((A\setminus B)\cup C = A\setminus (B\cup C)\) holds true.
Hence, the given conjecture \(17\) is proven to be correct for all sets \(A\), \(B\), and \(C\).