This fact does not cripple the proof. No binary turing decider of size x could run longer on an input of binary length n than x2^nBB(x), which is still an upper bound which can be used to solve the halting problem. (This is a tighter upper bound than that given by anon19 below, since BB(x) grows at a superexponential rate.)
This fact does not cripple the proof. No binary turing decider of size x could run longer on an input of binary length n than x2^nBB(x), which is still an upper bound which can be used to solve the halting problem. (This is a tighter upper bound than that given by anon19 below, since BB(x) grows at a superexponential rate.)