Special case of channel garbling/degradation

Let $P(X,Y)$ be a finite state distribution. Let’s build the Markov chain $X – Y – Y^1$ where $Y^1$ is a cover of $Y$ featuring only non-singleton subsets of $Y$ with probability $P(X, Y, Y^1)$ defined as
begin{equation}
p(x, y, y^1)
:=
begin{cases}
frac{p(x,y^1)}{r_y} & text{if } y in y^1 \
0 & text{otherwise}
end{cases}
end{equation}

I am finding that $P(Y^1 mid Y)$ is a degradation of $P(X mid Y)$, i.e. that
$$
p(y^1 mid y) = sum_x p(y^1 mid x) cdot p(x mid y)
quad
text{for all}
quad
y in Y
quad
text{and}
quad
y^1 in Y^1
$$

This result is confusing me, since we are not dealing with a chain $Y – X – Y^1$. The proof that I have drafted is below. Is the proof correct, or am I missing something?

Proof

From the relationship
$$
frac{p(x, y)}{p(y)}
=
frac{p(x, y mid y^1)}{p(y mid y^1)},
$$

we have
begin{equation}
p(y^1 mid y) = frac{p(x,y,y^1)}{p(x, y)}
=
begin{cases}
frac{1}{r^1_y} & text{if } y in y^1 \
0 & text{otherwise}
end{cases}
end{equation}

Furthermore we have
$$
sum_{x} p(y^1 mid x) cdot p(x mid y) =
sum_{x} frac{p(x,y^1) cdot p(x, y)}{p(x) cdot p(y)} =
begin{cases}
sum_{x} frac{p(x) cdot p(x, y)}{r^1_y cdot p(x) cdot p(y)} = frac{1}{r^1_y} & text{if } y in y^1 \
0 & text{otherwise}
end{cases}
$$