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So I am trying to show the following:

##(A \cup B)\sqcup(A \cap B) \leftrightarrow A \sqcup B##

The proof that I am trying to understand starts with:

##A \leftrightarrow (A \backslash B) \sqcup (A\cup B) \qquad (1)##,

and

##A \cup B \leftrightarrow (A\backslash B)\sqcup B \qquad (2)##

These two both make sense, in the first one we see that ##(A \backslash B) ## with ## (A\cup B)## is a partition of A, and thus there are bijections from ##(A \backslash B) ## to ## \{ (x,0) : x \in (A \backslash B) \}## and from ##A \cap B## to ##\{ (y,1) : y \in A \cap B \}##, and the same argument for ##(2)##

The next step in the proof says that:

##(A \cup B)\sqcup(A \cap B) \leftrightarrow (A\backslash B)\sqcup B \sqcup (A \cap B) ## which (apparently) follows from ##(2)##

and then, ##(1)## apparently means that

##(A\backslash B)\sqcup B \sqcup (A \cap B) \leftrightarrow A \sqcup B## which proves it.

But I don't understand how ##(1)## and ##(2)## helps us in these steps, can you just exchange terms like that? Anyone care to explain? Thanks..

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# Discrete mathematics, bijections between disjoint unions

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