For all sets A, B and C, if $(A \cap B) \cup C \subseteq A \cap (B \cup C) $, then $ C – (A \cup B) = \phi $.
I know this statement is true, as the antecedent implies that there is no element in C that is not also an element of A or B.
Furthermore, we can assume A, B, and C are sets and that $(A \cap B) \cup C \subseteq A \cap (B \cup C) $.
I believe we can prove this statement by contradiction, that is, by supposing $ C – (A \cup B) \ne \phi $ and showing how that leads to $(A \cap B) \cup C \not\subseteq A \cap (B \cup C) $, which would contradict our previous assumption that $(A \cap B) \cup C \subseteq A \cap (B \cup C) $.
…buuut I've been having trouble with the nitty gritty of the proof-by-contradiction. Any help would be greatly appreciated!
Best Answer
Continuing by contradiction as you have started:
So, suppose $C\setminus (A\cup B)\neq \emptyset$.
So, we have some $c$ such that $c\in C$ while $c\notin A$ and $c\notin B$.
Since $c\in C$ we have $c\in (A\cap B)\cup C$.
But, since $c\notin A$ we have that $c\notin A\cap (B\cup C)$, contradicting that $(A\cap B)\cup C$ should be a subset of $A\cap (B\cup C)$.