Why use the entanglement in Conway's free will theorem?

In Conway's Strong Free Will Theorem paper, the proof consists of two parts. In the first part, they proved the Specker-Kochen theorem. In the second part, they constructed two experimenters that are space-like separated, each possessing one spin-1 particle from an entangled pair, then they each made a measurement.

I don't see the point of making up the entanglement. From what I understood, the proof is like this:

• Suppose particles have no free will, but experimenters have, then there are two functions $$\theta_a(x, y, z), \theta_b(w)$$, such that $$\theta_a(x, y, z)\in \{(110),(101),(011)\}$$, and $$\theta_b(w)\in\{0, 1\}$$.
• By entanglement, we have $$\theta_a(x, y, z) = (\theta_b(x), \theta_b(y), \theta_b(z))$$.
• So $$\theta_b$$ satisfies the 101 property, which is impossible.

However, this just looks like the Specker-Kochen theorem with extra steps. Why is it necessary to show that $$\theta_b$$ has the 101 property by detouring through $$\theta_a$$? Is it supposed to squeeze out any possible contexuality?

Perhaps, phrased in another way, I'm asking this: suppose the universe has exactly one spin-1 particle, then is it possible that $$\theta_b$$ does not have the 101 property?

What it means in my notation is that, if we had only one spin-1 particle a, then the function $$\theta_a(x, y, z)$$ might not be of form $$\theta_a(x, y, z) = (\theta_b(x), \theta_b(y), \theta_b(z))$$, due to "contextuality", meaning that the squared-spin in direction x can be affected by the other two directions the experimenter will measure.
And as for why $$\theta_b$$ alone is not enough, I suppose it is also due to contextuality: if we only have $$\theta_b$$, we can't show that it has the 101 property without the help of $$\theta_a$$. Suppose we choose direction $$w$$, and measure the squared-spin of particle b along $$w$$. That gives $$\theta_b(w)$$, but now if we want to measure some other direction $$w'$$, we would be using $$\theta_b'(w')$$, which is not $$\theta_b$$, and thus we cannot show that $$\theta_b$$ itself has 101 property.