Return to Answer

I think the original question to be valid and Ciarán Taaffe's answer to be useful: Consider a diagonalisation on Turing machines, computing total computable functions from N to N. Then, G would be given a number, select the corresponding TM from the assumed enumeration, calculate the value and so on. But by assuming G was part of this list, there is a recursion (if G is given its own number in the enumeration) and therefore G does not terminate on that number and, in conclusion, does not compute a total function. With this contradiction, only the assumption that G was on that enumeration, is wrong (since it does not compute a total funtcion) and does not help proving more than that.

I think the original question to be valid and Ciarán Taaffe's answer to be useful: Consider a diagonalisation on Turing machines, computing total computable functions from N to N. Then, G would be given a number, select the corresponding TM from the assumed enumeration, calculate the value and so on. But by assuming G was part of this list, there is a recursion (if G is given its own number in the enumeration) and therefore G does not terminate on that number and, in conclusion, does not compute a total function. With this contradiction, only the assumption that G was on that enumeration, is wrong (especially since it would not compute a total function if it were) and does not help proving more than that.

I think the original question to be valid and Ciarán Taaffe's answer is useful: Consider a diagonalisation on Turing machines, computing total computable functions from N to N. Then, G would be given a number, select the corresponding TM from the assumed enumeration, calculate the value and so on. But by assuming G was part of this list, there is a recursion (if G is given its own number in the enumeration) and therefore G does not terminate on that number and, in conclusion, does not compute a total function. With this contradiction, only the assumption that G was on that enumeration, is wrong (since it does not compute a total funtcion) and does not help proving more than that.

I think the original question to be valid and Ciarán Taaffe's answer to be useful: Consider a diagonalisation on Turing machines, computing total computable functions from N to N. Then, G would be given a number, select the corresponding TM from the assumed enumeration, calculate the value and so on. But by assuming G was part of this list, there is a recursion (if G is given its own number in the enumeration) and therefore G does not terminate on that number and, in conclusion, does not compute a total function. With this contradiction, only the assumption that G was on that enumeration, is wrong (since it does not compute a total funtcion) and does not help proving more than that.

I think the original question to be valid and Ciarán Taaffe's answer is useful: Consider a diagonalisation on Turing machines, computing total computable functions from N to N. Then, G would be given a number, select the corresponding TM from the assumed enumeration, calculate the value and so on. But by assuming G was part of this list, there is a recursion (if G is given its own number in the enumeration) and therefore G does not terminate and, in conclusion, does not compute a total function. With this contradiction, only the assumption that G was on that enumeration, is wrong (since it does not compute a total funtcion) and does not help proving more than that.

I think the original question to be valid and Ciarán Taaffe's answer is useful: Consider a diagonalisation on Turing machines, computing total computable functions from N to N. Then, G would be given a number, select the corresponding TM from the assumed enumeration, calculate the value and so on. But by assuming G was part of this list, there is a recursion (if G is given its own number in the enumeration) and therefore G does not terminate on that number and, in conclusion, does not compute a total function. With this contradiction, only the assumption that G was on that enumeration, is wrong (since it does not compute a total funtcion) and does not help proving more than that.