Why are these two propositional expressions equivalent?

Question

We got this table here:

And from this table the following equation has been created (I understand so far and I understand how this equation has been read from the table!): $$c_{out}=(\overline{x}\land y \land c_{in})\lor (x\land \overline{y} \land c_{in})\lor (x\land y \land \overline{c_{in}})\lor (x\land y \land c_{in})$$

Here is the part I don't understand at all, how this equation was formed to: $$c_{out}=c_{in}\land (x\oplus y)\lor (x\land y)$$

Why we got only one $c_{in}$ in this formula and why there is this XOR sign, I don't understand anything : /

About this $c_{in}$ its because $c_{in}$ and $\bar{c_{in}}$ annule each other so there is $c_{in}$ left 2 times and instead of writing it 2 times we can write it 1 time. Is that correct or not?

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I have asked my professor and he just said "I'm on vacation", asked some other students and they couldn't help me too they just said I need to keep this formula in mind but that's bs you cannot keep everything in mind there must be a way to get there.

I hope someone can help me with this, don't know another way understanding it...

Answer 1

Hint: The expressions $(P\land Q)\lor (P\land R)$ and $P\land(Q\lor R) $ are logically equivalent for all $P,Q,R,$ so you can replace instances of the former with the latter in any propositional formula. Use this twice and simplify the resulting formulas.

Answer 2

It should be clear from the final formula that he has eliminated $\overline{c_{in}}$ . To do that take the 3rd and 4th terms together. Do you see it now. The final formula would be. $$ c_{out}= (c_{in}\land(x\oplus y)) \lor(x\land y)$$. Just to make it clearer. Do you see the derevation now?