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The problem statement is in the title.

I'm proving a problem in class and I need to show the above containment. I've drawn some Venn diagrams to make sure the containment makes sense, and it does to me, but I am still having problems proving this statement. I know that $A \Delta B = (A \cup B) – ( A \cap B)$, and I can see through the Venn diagrams I have drawn how [$C \subset A \Delta B$ iff $C \subset A \cup B$ and $A \cap B \cap C =\emptyset $], but I am having trouble explaining this to someone else.

Asaf Karagila
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Gabe Carr
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2 Answers2

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Let's try necessity. Let $$c \in C \subseteq A \Delta B = (A \cup B) - (A \cap B).$$ Then, $c \in A \cup B$ and $c \not \in A \cap B$. Thus,

  • every $c \in C$ also satisfies $c \in A \cup B$, hence, $C \subseteq A \cup B$;
  • every $c \in C$ also satisfies $c \not \in A \cap B$, hence $C \cap A \cap B = \emptyset$.

Can you go the other way?

gt6989b
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  • C ⊆ A Δ B =
    C ⊆ (A ⋃ B) ⋂ ¬(A ⋂ B) =
    C ⊆ (A ⋃ B) ⋂ (¬A ⋂ ¬B) =
    C ⊆ (A ⋂ ¬B) ⋃ (B ⋂ ¬A)
    Since c ϵ C also satisfies either A or B, but not A and B, the statement A ⋂ B ⋂ C must be an empty or null set. I’m still learning, so please tell me where I am either incorrect or where I could clean up my terminology.     – Gabe Carr Apr 09 '14 at 17:05
  • @GabeCarr not sure what you mean. The event $z \in X - Y = X \cap \bar{Y}$ happens if and only if $z \in X$ and $z \not \in Y$. – gt6989b Apr 09 '14 at 17:07
  • What I was trying to convey was that C is a subset of A or B, but that C is not a subset of A and B. – Gabe Carr Apr 09 '14 at 17:40
  • @GabeCarr the phrase "the statement A ⋂ B ⋂ C must be an empty or null set" is wrong: $A \cap B \cap C$ is not a statement, and statements cannot be empty or null sets. What is meant to say is that the set $A \cap B \cap C$ is empty, i.e. $A \cap B \cap C = \emptyset$ – gt6989b Apr 09 '14 at 17:44
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I would approach this as a simplification problem, which is straightforwardly solved by expanding the definitions from set theory and then simplifying using the rules of logic.

So I would start on the most complex side, and calculate as follows: \begin{align} & C \subseteq A \cup B \;\land\; A \cap B \cap C = \varnothing \\ \equiv & \qquad \text{"definitions of $\;\subseteq\;$; basic property of $\;\varnothing\;$"} \\ & \langle \forall x :: x \in C \;\Rightarrow\; x \in A \cup B \rangle \;\land\; \langle \forall x :: \lnot(x \in A \cap B \cap C) \rangle \\ \equiv & \qquad \text{"definition of $\;\cup\;$, and of $\;\cap\;$ twice; merge quantifications"} \\ & \langle \forall x :: (x \in C \;\Rightarrow\; x \in A \lor x \in B) \;\land\; \lnot(x \in A \land x \in B \land x \in C) \rangle \\ \equiv & \qquad \text{"logic: write $\;P \Rightarrow Q\;$ as $\;\lnot P \lor Q\;$ in left part;} \\ & \qquad \phantom{\text{"}}\text{DeMorgan for right part} \\ & \qquad \phantom{\text{"}}\text{-- to get everything in similar shape"} \\ & \langle \forall x :: (x \not\in C \lor x \in A \lor x \in B) \;\land\; (x \not\in A \lor x \not\in B \lor x \not\in C) \rangle \\ \equiv & \qquad \text{"logic: extract common disjunct $\;x \not\in C\;$"} \\ & \langle \forall x :: x \not\in C \lor ( (x \in A \lor x \in B) \;\land\; (x \not\in A \lor x \not\in B) ) \rangle \\ \equiv & \qquad \text{"logic: simplify $\;(P \lor Q) \land (\lnot P \lor \lnot Q)\;$ to $\;P \not\equiv Q\;$"} \\ & \langle \forall x :: x \in C \;\Rightarrow\; ( x \in A \not\equiv x \in B ) \rangle \\ \equiv & \qquad \text{"definition of $\;\triangle\;$; definition of $\;\subseteq\;$"} \\ & C \subseteq A \triangle B \\ \end{align} This completes the proof.

In the last step I used a not-too-well-known definition of set difference: $$ x \in A \triangle B \;\equiv\; x \in A \not\equiv x \in B $$

MarnixKlooster ReinstateMonica
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