For all sets $A$, $B$, $C$, if $C - (A \cup B) = \emptyset$ then $(A \cap B) \cup C \subseteq A \cap (B \cup C)$?

1. I need to prove or disprove the question in the title. I figured out the conclusion isn't true.

Since when I take $$A = \{ 1, 2 \}$$, $$B = \{ 2, 3 \}$$, and $$C = \{ 5 \}$$, $$(A \cap B) \cup C = \{ 2, 5 \}$$ and $$A \cap (B \cup C) = \{ 2 \}$$. So $$\{ 2, 5 \}$$ is not a subset of $$\{ 2 \}$$ and the conclusion of that statement is false. So is the statement as a whole true or false? If I had to prove it how would I prove it?

2. What if the statement was reversed, saying for all sets $$A$$, $$B$$, and $$C$$, if $$(A \cap B) \cup C \subseteq A \cap (B \cup C)$$, then $$C − (A \cup B) = \emptyset$$. Would that be true or false?

• I think i found a counter example to the first one proving its false. There exists sets A,B and C such that C − (A ∪ B) = ∅ but ( A ∩ B) ∪ C is not subset of A ∩ (B ∪ C) . If i take C = {1,2,3} A= {1,2} B= {2,3} . The negation of the statement appears to be true. So i can say the original statement is false. Am i right on this one or am i missing something? – Mahir Shahriar Oct 19 '18 at 14:33
• Your counterexample doesn’t satisfy hypothesis of 1. – Mayuresh L Oct 19 '18 at 14:35

Your attempt at 1 isn't correct, because $$C-(A\cup B)=\{5\}\ne\emptyset$$.
Saying that $$C-(A\cup B)=\emptyset$$ is the same as saying that $$C\subseteq A\cup B$$ On the other hand, $$A\cap B$$ could be empty. In this case the given inclusion would read $$C\subseteq A\cap(B\cup C)$$ If we take $$C=A\cup B$$, the main hypothesis is certainly satisfied. With these additional assumptions, $$A\cap(B\cup C)=A\cap C=A$$ and you just need to take $$B\ne\emptyset$$ to find an explicit counterexample. So $$A=\emptyset$$, $$B=C=\{1\}$$. Then $$(A\cap B)\cup C=\{1\}$$ but $$A\cap(B\cup C)=\emptyset$$.
If you don't like the empty set, take $$C=\{1,2\}$$, $$A=\{1\}$$, $$B=\{2\}$$. Then $$(A\cap B)\cup C=\{1,2\}$$ whereas $$A\cap(B\cup C)=\{1\}\cap\{1,2\}=\{1\}$$