Unconfounded assumption

In the notation of the unconfounded assumption, does $$\left(Y(0),Y(1)\right)\perp W \mid X$$

mean

$$f(Y(0),Y(1), W\mid X)=f(Y(0),Y(1)\mid X)\cdot f(W\mid X)$$ ?

I can prove that the second line if the set of random variables $(Y(0),Y(1))$ is independent of $W$ given $X$: $$f(Y(0),Y(1), W\mid X)=f(Y(0),Y(1)\mid W,X)\cdot f(W|X)$$ by the conditioning rule. Since $f((Y(0),Y(1)|W,X)$ does not depend on $W$ by the assumption stated, the result is obtained. But every signle paper omits this discussion. I am studying this treatment effect literature by myself, so I need to understabd this fundamental assumption based on my econometrics knowledge.

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The notation $Y \perp W$ means that the random variable $Y$ is statistically independent of $W$. So, assuming $(Y,W)$ has a joint density $f_{Y,W}(y,w)$, we have $$f_{Y,W}(y,w) = f_Y(y) f_W(w).$$ Similarly, the notation $Y \perp W \mid X$ means that, conditional on $X$, $Y$ is statistically independent of $W$. Again assuming that all random variables have densities, we have $$f_{Y,W \mid X}(y,w \mid x) = f_{Y \mid X}(y \mid x) f_{W \mid X}(w \mid x).$$ The results above hold if $Y$, $W$, and $X$ are random vectors. So, the answer to your question is 'yes'. Papers and textbooks which discuss unconfoundedness and treatment effects assume you know the definition of statistical independence, which is why they don't discuss this. You may want to spend some time reviewing probability theory before continuing your self study of treatment effects.