If service A feeds both B and C, and they both feed service D, then D can receive an incoherent view of what A did, because nothing forces B and C to keep their stories straight. But B and C can still both be following their own spec perfectly, so there's no bug in any single service. Now it's not clear whose job it is to fix things.

I wouldn't structure a team that way now.

All the (non-limit) von Neumann ordinals are of the form X+1 = {X, {X}}, where X is the previous ordinal in the set. If you just look at trees of this form:

X+1: X <- node -> {X}, or X <- node -> node -> X

then you ignore the direction of the parent-child relation, you get this:

X+1: X -- node -- node -- X

So that's why your trees are symmetric as undirected graphs; and of course, every lower ordinal has its own version of this symmetry, which is also contained in the tree. All the large gaps between sections correspond to node--node edges of the larger ordinals. Kinda neat!

Hah! I agree with this. :)

Fermat: x^N + y^N = z^N. The Diophantine set only has the number N=2, because Fermat's Last Theorem tells us that no other possible N will work if x, y, z have to be integers. This is just a finite set.

Other polynomial expressions work with an infinite number of parameter settings: Pell's equation is x^2 - N*y^2 = 1. In this case, every N that is not a square will work. This is an infinite set.

Because a lot of the parameter sets are infinite like this, the size of the parameter set are instead measured using big-O notation. For Fermat's Last Theorem, {N=2} is constant, so big-O(1). For Pell's equation, {N not a square} is O(N). For other Diophantine expressions, the parameter settings are O(N^2), or O(2^N), or any other growth function you want that's achievable with Turing machines.

And this is the essence of what MRDP proved. Robinson, Davis, Putnam, and Matiyasevich over several papers proved that if you could encode a set that grew roughly like O(2^N), you also could encode any other computable set in the parameters. Approximately, demonstrating an exponentially-fast-growing family of parameters for some specific Diophantine problem was enough for Diophantine problems as a whole to be Turing complete. Matiyasevich's clinching discovery was a concrete example using the Fibonacci numbers.

Collatz: if you replace 1 and 3 with arbitrary parameters, Conway showed that this class of problems is undecidable, so any particular instance could be arbitrarily hard.