However, it seems fairly clear that the corresponding calculation with three generators is too large to be done by hand, and I can't see any "pattern" emerging from the two-generator one which might indicate why they should all be finite.
I see a pattern that very strongly inclines me to the conjecture that they are all finite. Peter's poset is simply the product of three chains, 2 x 3 x 3. One rarely sees that kind of regularity in a combinatorial enumeration that blows up to infinity at some finite point. The one-generator chain being simply 2 (1 and x), the question then becomes, what's the next element in the sequence 2, 2 x 3 x 3, ...? I'd like to see one more term before going out on a limb here. :) Vaughan Pratt 12-Jan-2005 12:21:18 -0400,2004;000000000000-00000000