Given four sets: $A, B, C, D$, suppose that $$A \triangle B \subseteq D \ \land \ B \triangle C \subseteq D $$
**Prove:** $A \triangle C \subseteq D$

[As $\triangle$ means symmetric difference] 

***My Attempt:*** 

Notice that $A \triangle B = (A \setminus B) \cup (B \setminus A) = (A \cup B)\setminus (A\cap B)$ 

Therefore: $\left((A\cup B)\setminus(A\cap B)\right) \text{ and } \left((B\cup C)\setminus (B\cap C)\right) \subseteq D$

***EDIT:***
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I'd like to prove: $(A \triangle C) \subseteq (A\triangle B)\cup(B\triangle C)$
$$\left((A\triangle B)\cup(B \triangle C)\right)=(A\cup B)\setminus(A\cap B)\cup(B\cup C)\setminus(B\cap C)$$ $$ =(A\cup B)\cup(B\cup C)\setminus(A\cap B)\cup(B\cap C)$$
$$=B\cup(A\cup C)\setminus B\cap(A\cup C)$$

$ \ $

**Now** -- can I say that $(A\cup C)\setminus(A\cap C) \subseteq (B\cup(A\cup C)\setminus B\cap(A\cup C)$?

***my previous attempt:***
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Notice that $A \triangle B = (A \setminus B) \cup (B \setminus A) = (A \cup B)\setminus (A\cap B)$ 

Therefore: $\left((A\cup B)\setminus(A\cap B)\right) \text{ and } \left((B\cup C)\setminus (B\cap C)\right) \subseteq D$

**I'll divide into cases:** 

**I.** let $x \in (A\triangle B) \land x\notin C \to x\in (A\setminus C) \subseteq (A \triangle C)$ 

Notice that $A\triangle B \subseteq D \to x\in D$, therefore $x \in A\triangle C \land x\in D$ 

**II.** let $x'\in (B \triangle C) \land \ x'\notin A \ $, then $x'\in (C\setminus A) \subseteq (A\triangle C)$ **as** $(B\triangle C) \subseteq D \to \ x'\in D$ 

**III.** Let $x'' \in (B\cap C) \land x\notin A$ Then $x\in (B\setminus 
A)\subseteq (A \triangle B) \subseteq D$, Hence $x'' \in D$

**Notice that** $x'' \in (C\setminus A) \subseteq (A \triangle C)$ , so $x'' \in D \land x''\in (A\triangle C)$. 

*I don't know if I covered all possible cases (and can't find a way to do so. using contradiction didn't helped).
What can be done at such case?*