Something about that proof feels wrong.. Are you sure the allusion to the halting problem applies here? The proof that the halting problem can't be computed r

Something about that proof feels wrong.. Are you sure the allusion to the halting problem applies here? The proof that the halting problem can't be computed relies on a self-reference where you query whether the program halted and then do the opposite. But- and here's the important part- when the program references itself it requires at least one more symbol to do so as it needs somewhere to store the answer. It therefore needs an infinite stack. But we have a fixed number of symbols. Or, to put it another way: if your set of symbols is bounded, then you can only call functions that use at least one fewer symbol than the calling one, which means you will eventually run out of symbols and have a base case. Doesn't that mean the halting problem doesn't apply to the busy beaver problem?