Following the example of the 1st equation, why doesn't the 2nd and 3rd equation result in
$$\sum_{all\,x} \,\, \sum_{y:y=x} P(x,y) $$
$$\sum_{all\,x} \,\, \sum_{y:y=z-x} P(x,y) $$ ?
2026-03-25 20:35:21.1774470921
How did these equations go from Step 2 to Step 3?
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It does. However $$\sum_{y\colon y=z}f(y)=f(z).$$