Suppose $A,B,C$ are sets. Prove that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$.

Solution 1:

Lets have our three sets:

We find $A\mathbin\triangle B$ as the elements in $A$ or in $B$ but not in both. corresponding to the yellow area bellow:

Given that $A\mathbin\triangle B\subseteq C$, then the red area bellow is empty:

So if we take $A\cup C$ minus the empty area (red above) we will get the green area below:

Which is the same result as if we get $B\cup C$, which proves: $A\mathbin\triangle B\subseteq C\implies A\cup C=B\cup C$.

Now for the other way:

If $A\cup C=B\cup C$ then the red areas bellow must be empty:

When we substract these empty areas from the yellow areas from $A\mathbin\triangle B$ (bellow):

then the reminder is contained in $C$.

This proves that $A\cup C=B\cup C\implies A\mathbin\triangle B\subseteq C$.

Solution 2:

I'll show that every $x\in A\cup C$ is also in $B\cup C$, other way is symmetrical. Assume $x\in A\cup C$. If $x\in C$ then it is in $B\cup C$. Then assume $x\in A$. If $x\in A$ but not in $B$ then $x\in C$ from $A\Delta B \subseteq C$. Then assume $x\in A\cap B$. Then $x\in B$ and therefore $x\in B\cup C$. $\square $