> This clearly solves the halting problem (we assumed the existence of halt as the first step of our proof by contradiction).
Implementing the halting problem using addition does not assume the existence of halt. It assumes the existence of addition, and you use that to show that halt must also exist. And that's not possible to do.
To implement halts using addition, you need to:
- Take the input to halts, namely the source code to a program.
- Convert that into an input to addition, namely two numbers.
- Call into addition, which adds the numbers.
- Use the result of that addition to infer whether the original program halts or not.
Why are you debating the validity of the argument?
The code very clearly:
- is an implementation of the halting problem. If halt existed it would undoubtly solve the halting problem
- it uses '+'
You are prentending I didn't know what the author meant. The point is that what the author says is simply not true, if you take the statement as written.
What he wants to show is that if canBeRegex is computable then halt is computable, but that isn't what he is saying.
Implementing the halting problem using addition does not assume the existence of halt. It assumes the existence of addition, and you use that to show that halt must also exist. And that's not possible to do.
To implement halts using addition, you need to:
- Take the input to halts, namely the source code to a program.
- Convert that into an input to addition, namely two numbers.
- Call into addition, which adds the numbers.
- Use the result of that addition to infer whether the original program halts or not.