# $A, B,$ and $C$ are all sets that lie in a common universal set. Prove $A = B$ given the following statements.

$$(A \cup C) = (B \cup C)$$ and $$(A \cap C) = (B \cap C)$$

I know I have to start with $$A = ...$$ to get to $$B$$. I incorporated $$A = A \cap (A \cup C)$$ as the first step using absorption law, but I don't know where to go from here. How do I show that it equals $$(A \cup C) = (B \cup C)$$? I've looked at the set identities, and think that I have to use absorption law again to show that $$A \cap (A \cup C) = (A \cap (A \cup C)) \cap ((A \cap (A \cup C)) \cup (B \cup C))$$. Can you guys please help me solve this problem, I'm stuck.

If the element is in $$A$$ then either it's in $$C$$ or it isn't.

If it is in $$C$$ then it is in their intersection, which implies that it is in the intersection of $$B$$ and $$C$$ which implies that it is in $$B$$.

If it is not in $$C$$, it is still in the union of $$A$$ and $$C$$, which implies that it is in the union of $$B$$ and $$C$$, which implies that it is in $$B$$ (since it is assumed here that it is not in $$C$$).

By a symmetrical argument, $$B$$ is also a subset of $$A$$.

You can use absorption and distributivity: \begin{align} A&=A\cap(A\cup C)&&\text{absorption}\\ &=A\cap(B\cup C)&&\text{assumption}\\ &=(A\cap B)\cup(A\cap C)&&\text{distributivity}\\ &=(A\cap B)\cup(B\cap C)&&\text{assumption}\\ &=(B\cap A)\cup(B\cap C)&&\text{commutativity}\\ &=B\cap(A\cup C)&&\text{distributivity}\\ &=B\cap(B\cup C)&&\text{assumption}\\ &=B&&\text{absorption} \end{align} You can see that this holds in every distributive lattice.

Let me try a direct approach. For sets $$A$$ and $$C$$, we have one of the following two possibilities:

1. $$A \subset C$$, this includes the case where both sets are equal;

2. $$A \cap C$$ is some proper subset of $$A$$, which may be empty.

For the first case, we know that $$A \cup C = C = B \cup C$$, hence $$B \subset C$$. By the second equality given in the question, we have $$A = A \cap C = B \cap C = B$$.

For the second case, let $$A \cap C = D = B \cap C$$. By the first equality in the question, $$(A \cup C) \backslash C = (B \cup C) \backslash C \implies A \backslash D = B \backslash D$$. Taking union with $$D$$ on both sides gives us the desired result.

I suggest a Venn diagram of $$A, B, C$$.

First solution.

Let $$S_1=A$$ \ $$(B\cup C)$$ and $$S_2=B$$ \ $$(A\cup C)$$ and $$S_3=C$$ \ $$(A\cup B).$$

Let $$S_4=(A\cap B)$$ \ $$C$$ and $$S_5=(A\cap C)$$ \ $$B$$ and $$S_6=(B\cap C)$$ \ $$A .$$

Let $$S_7=A\cap B\cap C.$$

Then $$\{S_j:1\le j\le 7\}$$ is a pairwise-disjoint family.

For brevity let $$T=\cup_{j=3}^{j=7}\,S_j.$$ Then $$S_1\cup T=A\cup C=B\cup C=S_2\cup T.$$ Therefore $$S_1=S_1\cap (S_1\cup T)=S_1\cap (S_2\cup T)=\emptyset.$$ Interchanging the subscripts $$1, 2$$ in the preceding line, we also have $$S_2=\emptyset.$$

We have $$S_5=S_5\cap ( S_5\cup S_7)=S_5\cap (A\cap C)=S_5\cap (B\cap C)=S_5\cap (S_6\cup S_7)=\emptyset.$$ And similarly we obtain $$S_6=\emptyset.$$

Therefore $$A=S_1\cup S_5\cup S_4\cup S_7=S_2\cup S_6\cup S_4\cup S_7 =B.$$

Second solution.

For any $$A,B,C$$ we have $$A=(A\cap C)\cup (A\setminus C)=(A\cap C)\cup ([A\cup C]\setminus C)$$ and $$B=(B\cap C)\cup (B\setminus C)=(B\cap C)\cup ([B\cup C]\setminus C)$$. Now compare the RHS of these if $$A\cap C =B\cap C$$ and $$A\cup C=B\cup C.$$

• I have added a second solution which is shorter and simpler then my first one. Commented Mar 1, 2021 at 20:24

Assuming $$(A \cup C) = (B \cup C)$$ and $$(A \cap C) = (B \cap C)$$. Prove $$A=B$$.

Assume $$x \in A$$ and $$x \notin B$$.
Then $$x \in B \cup C \to x \in$$ C.
But this leads to a contradiction. Because $$x \notin B \cap C$$ but $$x \in A \cap C$$.
Therefore, if $$x \in A$$ then $$x \in B$$.
And the same argument works for $$x \in B$$ but $$x \notin A$$.
Therefore, if something is in $$A$$ then it is in $$B$$ and vice versa.