I am reading through the Fair ML Book by Barocas et al. In chapter 4, they state that for all values $a,b$ that random variable $A$ can take on
$$P[Y = 1 | A = a, R = r] = P[Y = 1 | A = b, R = r]$$
is a consequence of sufficiency (or $Y \perp A | R$) when $Y \in \{0, 1\}$.
I'm trying to derive the equality from $Y \perp A | R$ and $Y \in \{0, 1\}$ by applying the definition of independence and trying to substitute. However that hasn't really lead anywhere as far as I can tell.
Note: The definition for independence of random variables is a universal statement.
For discrete random variables $Y,A,$ and $R$, we have that $Y\perp A\mid R$ means exactly that:
$$\forall y\forall x\forall z\quad\mathsf P(Y{=}y, A{=}x\mid R{=}z)=\mathsf P(Y{=}y\mid R{=}z)\,\mathsf P(A{=}x\mid R{=}z)$$
So by definition of conditional probability:
$$\forall y\forall x\forall z\quad\big(\mathsf P(A{=}x\mid R{=}z)\neq 0\to \mathsf P(Y{=}y\mid A{=}x, R{=}z)=\mathsf P(Y{=}y\mid R{=}z)\big)$$
Therefore for arbitrary values, $a,b,r$, as long as $\mathsf P(A{=}a\mid R{=}r)\neq 0$ and $\mathsf P(A{=}b\mid R{=}r)\neq 0$, we have:
$$\mathsf P(Y{=}1\mid A{=}a, R{=}r)=\mathsf P(Y{=}1\mid R{=}r) \\\text{and}\\ \mathsf P(Y{=}1\mid A{=}b, R{=}r)=\mathsf P(Y{=}1\mid R{=}r)$$
Therefore... for all $a, b,$ and $r$ (with the aforementioned restrictions)...
$$\mathsf P(Y{=}1\mid A{=}a, R{=}r)~=~\mathsf P(Y{=}1\mid A{=}b, R{=}r)$$