3 ms·

I am at a total loss here. Do you not understand that it is the point that my reasoning is invalid? Why are you pretending I do not understand the basics of lo

I am at a total loss here. Do you not understand that it is the point that my reasoning is invalid? Why are you pretending I do not understand the basics of logic while missing the actual argument by a mile?

Gehinnn 4y ago
Your reasoning is valid, and I do have the impression that you should refresh your knowledge about logic.
- constantcrying 4y ago
  As a reminder: a proof by contradiction works by assuming first the negation of a given statement. If from that you can deduce a falsehood you know that statement is true. In the article the author says: "If you implement halts using canBeRegex, you’ve got a proof by contradiction that canBeRegex can’t exist." Which is plainly a false statement. What he means is the following statement: "If you show that if canBeRegex is computable then halts is computable, you’ve got a proof by contradiction that canBeRegex can’t exist." Those two statements are clearly not equivalent. And only the second one is true.
  - dellamonica 4y ago
    Wait, so you are saying that the only thing missing in the first statement is that the implementation is required to be correct? I think that is pretty obvious. This is beyond nitpicking.
    - constantcrying 4y ago
      >that the implementation is required to be correct? No. "Implemented" is in fact irrelevant, it is clearly not the term to use here. Because what the implementation uses is irrelevant, the question is about whether computability in one can be derived from computability of the other. In his example the implementation also uses ".length", whether it does or doesn't is irrelevant though. Since it is not a question of implementation. >This is beyond nitpicking. Also known as standard practice in mathematics.
      - dellamonica 4y ago
        Implemented in this context means exactly the type of reduction that is needed. If halt can be implemented by using X that means you can reduce X to halt. This is similar to polynomial time reductions in complexity. Standard practice in mathematics would apply here if what was said is wrong. This time it is about an interpretation that no one shares with you. [edit] After reading again your comment I realized that maybe you are not applying the logic correctly. If the reduction "implementation" uses both length and foo, then the conclusion is that at least one of length or foo cannot be computable. Since length is trivial, then it must be foo that is not computable.
        constantcrying 4y ago
        >If halt can be implemented by using X that means you can reduce X to halt. Replace X with .length and read your statement again. "If halt can be implemented by using .length that means you can reduce .length to halt." Please just answer whether you believe that statement is true or false (or undecidable).
        dellamonica 4y ago
        You would need to show how to use length to implement halt and that is the part you will not be able to do. If you did, then yes, length would not be computable. Reductions in complexity theory work like this. An explicit algorithm invokes a black box and this shows that either the black box does not exist or you can solve the original problem. Typically you care about the complexity of the reduction to establish that the problem is inside a given complexity class (e.g. NP complete).
        constantcrying 4y ago
        You didn't answer the question. The statement is obviously false by the way, as you can see in the article where the author does exactly that. You rephrasing the statement to make it true, is just ignoring the point. I would encourage you to engage in some studying of reading comprehension. It is an important skill in mathematics and related disciplines.

Gehinnn 4y ago

Your reasoning is valid, and I do have the impression that you should refresh your knowledge about logic.

constantcrying 4y ago
As a reminder: a proof by contradiction works by assuming first the negation of a given statement. If from that you can deduce a falsehood you know that statement is true. In the article the author says: "If you implement halts using canBeRegex, you’ve got a proof by contradiction that canBeRegex can’t exist." Which is plainly a false statement. What he means is the following statement: "If you show that if canBeRegex is computable then halts is computable, you’ve got a proof by contradiction that canBeRegex can’t exist." Those two statements are clearly not equivalent. And only the second one is true.
- dellamonica 4y ago
  Wait, so you are saying that the only thing missing in the first statement is that the implementation is required to be correct? I think that is pretty obvious. This is beyond nitpicking.
  - constantcrying 4y ago
    >that the implementation is required to be correct? No. "Implemented" is in fact irrelevant, it is clearly not the term to use here. Because what the implementation uses is irrelevant, the question is about whether computability in one can be derived from computability of the other. In his example the implementation also uses ".length", whether it does or doesn't is irrelevant though. Since it is not a question of implementation. >This is beyond nitpicking. Also known as standard practice in mathematics.
    - dellamonica 4y ago
      Implemented in this context means exactly the type of reduction that is needed. If halt can be implemented by using X that means you can reduce X to halt. This is similar to polynomial time reductions in complexity. Standard practice in mathematics would apply here if what was said is wrong. This time it is about an interpretation that no one shares with you. [edit] After reading again your comment I realized that maybe you are not applying the logic correctly. If the reduction "implementation" uses both length and foo, then the conclusion is that at least one of length or foo cannot be computable. Since length is trivial, then it must be foo that is not computable.
      - constantcrying 4y ago
        >If halt can be implemented by using X that means you can reduce X to halt. Replace X with .length and read your statement again. "If halt can be implemented by using .length that means you can reduce .length to halt." Please just answer whether you believe that statement is true or false (or undecidable).
        dellamonica 4y ago
        You would need to show how to use length to implement halt and that is the part you will not be able to do. If you did, then yes, length would not be computable. Reductions in complexity theory work like this. An explicit algorithm invokes a black box and this shows that either the black box does not exist or you can solve the original problem. Typically you care about the complexity of the reduction to establish that the problem is inside a given complexity class (e.g. NP complete).
        constantcrying 4y ago
        You didn't answer the question. The statement is obviously false by the way, as you can see in the article where the author does exactly that. You rephrasing the statement to make it true, is just ignoring the point. I would encourage you to engage in some studying of reading comprehension. It is an important skill in mathematics and related disciplines.