Final answer:
Proof established that for non-empty sets A and B, where A is a subset of B, the intersection of B and the complement of A is equal to the set difference of B and A. This conclusion is derived by showing every element's presence in one set implies its presence in the corresponding set, confirming the equality B∩ (Ac) = B \ A.
Step-by-step explanation:
To show that if A and B are non-empty sets such that A ⊆ B, then B∩ (Ac) = B \ A, we need to prove that every element of B∩ (Ac) is in B \ A, and every element of B \ A is in B∩ (Ac).
Part 1: B∩ (Ac) ⊆ B \ A
Let x be an element of B∩ (Ac). This means that x is in B and not in A (since it's in the complement of A, Ac). Therefore, by the definition of set difference, x is in B \ A.
Part 2: B \ A ⊆ B∩ (Ac)
Now let x be an element of B \ A. This means that x is in B but not in A. Because x is not in A, it must be in the complement of A (Ac). Thus, x is also in B∩ (Ac).
Since every element of B∩ (Ac) is shown to be in B \ A, and every element of B \ A is shown to be in B∩ (Ac), we have proven that B∩ (Ac) = B \ A.