The “original Turing machine” decides whether a prefix is valid? Then yes definitely; bear in mind Rice’s theorem, which basically says that no non-trivial property of a Turing machine is computable.
The “original Turing machine” decides whether a prefix is valid?
If I understand you, yes. By the “original Turing machine”, I meant the machine T that putatively can interpret an input string as an integer or, alternatively, reject the input string as not corresponding to any integer.
So, can we actually construct such a machine that is provably immune to the “attack” in your proof, in the sense that no Turing machine could implement the attack? Or are you saying that Rice’s theorem (with which I will have to acquaint myself) says that pretty much any Turing machine T that maps strings to integers will fit the bill? (Though, the one in cousin_it’s OP doesn’t . . .)
Hmm. So T is an acceptor, A is an attacker. A(T) is an infinite sequence of symbols produced after examining T’s source code, and T(A(T)) = ⊥ means T never rejects the sequence. Then what I was asserting is ¬∃A:∀T:T(A(T)) = ⊥. What you’re asking now is whether ∀T:∃A:T(A(T)) = ⊥ and I can’t immediately work out the answer.
The “original Turing machine” decides whether a prefix is valid? Then yes definitely; bear in mind Rice’s theorem, which basically says that no non-trivial property of a Turing machine is computable.
If I understand you, yes. By the “original Turing machine”, I meant the machine T that putatively can interpret an input string as an integer or, alternatively, reject the input string as not corresponding to any integer.
So, can we actually construct such a machine that is provably immune to the “attack” in your proof, in the sense that no Turing machine could implement the attack? Or are you saying that Rice’s theorem (with which I will have to acquaint myself) says that pretty much any Turing machine T that maps strings to integers will fit the bill? (Though, the one in cousin_it’s OP doesn’t . . .)
Hmm. So T is an acceptor, A is an attacker. A(T) is an infinite sequence of symbols produced after examining T’s source code, and T(A(T)) = ⊥ means T never rejects the sequence. Then what I was asserting is ¬∃A:∀T:T(A(T)) = ⊥. What you’re asking now is whether ∀T:∃A:T(A(T)) = ⊥ and I can’t immediately work out the answer.