Cone of Gray code function

In recitation 8 it was presented that for gray code:
\mathrm{cone}\left( g[i] \right) = \left\{ b[j] \Big| i < j \le n \right\}
But for gray code it can be shown that:
g[n - 1] = b[n - 1]
g[i] = \mathrm{XOR}\left( b[i + 1],b[i] \right) \quad \left( \forall 0 \le i \le n - 2 \right)
So according to that isn’t \mathrm{cone}\left( g[i] \right) = \left\{ b[i + 1] , b[i] \right\} \quad \left( \forall 0 \le i \le n - 2 \right) only?

one question per post please.
add an informative title too.

You are correct, the cone of the function implemented by g[I] is \left\{ b[i], b[i + 1] \right\}.
The last line in the recitation slides is wrong, but the reasoning is correct. Following the reasoning will give you the same result.