I understand that, what I am saying is, that the upper bound can never be useful because the lower bound is already so high that we cannot run U+1 steps, ever.
I see; thanks for clarifying. I suppose the only thing you’d get “for free” is that the termination of these programs becomes decidable. (Not sure if this is known for these specific programs. At some point, BB number bounds are necessarily unknowable.)
_alternator_|7 months ago