Thanks! That was indeed useful. Actually, I did write to Shanon Fitzpatrick but haven't received any reply yet.

@MoisheKohan Thanks. Could you please provide an example, or a reference?.

@JosephO'Rourke Yes, I know (which is why I said just saying a graph has 2n-3 edges isn't enough to guarantee rigidity). But this is for generic configuration only. I am interested in showing rigidity for non-generic configuration.

@caduk Thanks for the reference and the example. But it does not give any argument (which is what I am interested in) to show that it is actually rigid.

@IlyaBogdanov Ah! I see, Thanks. I think that works. So basically, we start with a node $v$, then fold the edges incident to $v$ to a single edge $vv_1$, then fold the rest of the edges incident to $v_1$ to an edge $v_1v_2$, and so on. And this works for any bipartite graph with equal edge lengths (need not be skeleton graph of a polytope)

@IlyaBogdanov Yes continuous folding (the usual physical folding, i.e. the edges can not cross during folding). Could you please elaborate why it is obvious?

@RichardStanley: Thanks. I believe your claim is that at least $k$ chains/antichains are required to cover the poset you mentioned. Could you please provide an argument to prove this?