What about the other two proofs I've provided? I'm definitely fairly confident about the one in B02/S.
I may have a partial proof of impossibility for S345678 without B2, but the case analysis is too involved. The conjectured solution is that no pattern can expand too far out of the bounding diamond without forming an immortal polyplet; in the attatched example, it forms in generation 1 and is the hexomino occurring as generation 1 of prepond.
Code: Select all
x = 4, y = 4, rule = B3/S345678History
B8D$2B7D$3B6D$2B2A5D$3B2A4D$5BA3D$4B2AB2D$8BD$9B!
You may want to sketch this out.
First, if all the cells in a polyplet have 3 or more neighbors, then none of those cells can ever die. Therefore, any moving pattern must have an arrangement of cells that contains no such polyplet. I also note that when a pattern consists entirely of such an object, its evolution becomes identical to Life without Death.
Second, David Eppstein proved that with S234567 and without B2, no spaceships can exist because no pattern can escape its bounding diamond without containing an immortal triangle. Therefore, we will always assume that S2 is off.
Let (x, y) be a live cell that maximizes x+y. We want at least one of (x+1, y) or (x, y+1) to be live in generation 1. The only way for (x, y+1) to be born in generation 1 for (x-1, y) and (x-1, y+1) to be on, and similarly (x+1, y-1) and (x, y-1) must be on for (x+1, y) to be live in generation 1.
If both (x+1, y) and (x, y+1) are on in generation 1, then the five live cells mentioned so far form a "W" shape. But this means that (x-1, y) and (x, y-1) each have at least 3 neighbors and (x, y) has at least 4, so they survive to generation 1 and, together with (x, y+1) and (x+1, y), form an immortal "+" shape where each of these five cells always has at least three neighbors.
Then assume, without loss of generality, that only (x, y+1) is on in generation 1. We can then consider the remaining three possible live neighbors of (x, y). Note that (x-2, y+1) and (x-2, y+2) cannot both be on, or else an immortal + forms around (x-1, y+1). Nor can (x-1, y-1) and (x, y-1) both be on, or else we already have an immortal block. [What happens next?]
First, if (x-1, y-1) is on and (x, y-1) is off, then (x-1, y) and (x, y) are still on in generation 1. None of (x-2, y+2), (x-2, y-1), or (x-2, y) can be on, or else (x-1, y+1) will survive and we get an immortal block. Then (x, y+1) has 2 live neighbors and dies, so (x+1, y+1) has no more than two live neighbors in generation 2.
If (x+1, y-1) is on and (x-1, y-1) is off, then (x+1, y) is off in generation 1 and (x, y) and (x, y+1) are on. (x-2, y+2) and (x-2, y+1) cannot both be on because that makes a W and leads to an immortal +. If (x-2, y+1) or (x-2, y) are both on, then both (x-1, y+1) and (x-1, y) survive and an immortal block is formed. This also happens if (x-2, y+2) and (x-2, y-1) are both on. Therefore, the initial configuration must have only at most one of the latter two, in addition the already-assumed cells. [Again, what happens next?]