How to get the last function in the picture by o.1 and o.2 equations? (Conditional distribution)

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[for the last equation, is it shows that the sum of P(x given y) times p(y)? And i do not sure how this equation form by o.1 and o.2, please give me a help. Best wishes 1,

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By equation $0.1$, we have

$$P(X=x)=\sum_{y}P(X=x,Y=y)$$

By equation $0.2$, we have

$$P(X=x|Y=y)=\frac{P(X=x,Y=y)}{P(Y=y)}$$

and can rearrange to get

$$P(X=x|Y=y)P(Y=y)=P(X=x,Y=y)$$

Substitue the LHS for $P(X=x,Y=y)$ in the first line to get

$$P(X=x)=\sum_{y}P(X=x|Y=y)P(Y=y)$$