Show that $lambda x.(left {x right }(x)neq 0)$ is not recursive.
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Show that $lambda x.(left {x right }(x)neq 0)$ is not recursive.
I am trying to show this relation is not recursive using a code number for this relation so that I can ultimately prove that the decision problem for $lambda zx.(left {z right }(x)=0)$ is recursively unsolvable.
If I let $e$ be a code number for $lambda x.(left {x right }(x)neq 0)$, does this form a contradiction by assuming $g(x) = left {x right }(x)$ is a total recursive function? I am not sure how to use code numbers for this kind of relation.
logic recursion computability
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add a comment |
$begingroup$
Show that $lambda x.(left {x right }(x)neq 0)$ is not recursive.
I am trying to show this relation is not recursive using a code number for this relation so that I can ultimately prove that the decision problem for $lambda zx.(left {z right }(x)=0)$ is recursively unsolvable.
If I let $e$ be a code number for $lambda x.(left {x right }(x)neq 0)$, does this form a contradiction by assuming $g(x) = left {x right }(x)$ is a total recursive function? I am not sure how to use code numbers for this kind of relation.
logic recursion computability
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1
$begingroup$
You have to clarify whether ${x}(x) neq 0$ means (1) the computation ${x}(x)$ terminates and returns a non-zero result or (2) the computation ${x}(x)$ does not terminate with a zero result. With either reading, you can reduce the halting problem to this one.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:07
add a comment |
$begingroup$
Show that $lambda x.(left {x right }(x)neq 0)$ is not recursive.
I am trying to show this relation is not recursive using a code number for this relation so that I can ultimately prove that the decision problem for $lambda zx.(left {z right }(x)=0)$ is recursively unsolvable.
If I let $e$ be a code number for $lambda x.(left {x right }(x)neq 0)$, does this form a contradiction by assuming $g(x) = left {x right }(x)$ is a total recursive function? I am not sure how to use code numbers for this kind of relation.
logic recursion computability
$endgroup$
Show that $lambda x.(left {x right }(x)neq 0)$ is not recursive.
I am trying to show this relation is not recursive using a code number for this relation so that I can ultimately prove that the decision problem for $lambda zx.(left {z right }(x)=0)$ is recursively unsolvable.
If I let $e$ be a code number for $lambda x.(left {x right }(x)neq 0)$, does this form a contradiction by assuming $g(x) = left {x right }(x)$ is a total recursive function? I am not sure how to use code numbers for this kind of relation.
logic recursion computability
logic recursion computability
asked Dec 5 '18 at 1:36
numericalorangenumericalorange
1,745311
1,745311
1
$begingroup$
You have to clarify whether ${x}(x) neq 0$ means (1) the computation ${x}(x)$ terminates and returns a non-zero result or (2) the computation ${x}(x)$ does not terminate with a zero result. With either reading, you can reduce the halting problem to this one.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:07
add a comment |
1
$begingroup$
You have to clarify whether ${x}(x) neq 0$ means (1) the computation ${x}(x)$ terminates and returns a non-zero result or (2) the computation ${x}(x)$ does not terminate with a zero result. With either reading, you can reduce the halting problem to this one.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:07
1
1
$begingroup$
You have to clarify whether ${x}(x) neq 0$ means (1) the computation ${x}(x)$ terminates and returns a non-zero result or (2) the computation ${x}(x)$ does not terminate with a zero result. With either reading, you can reduce the halting problem to this one.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:07
$begingroup$
You have to clarify whether ${x}(x) neq 0$ means (1) the computation ${x}(x)$ terminates and returns a non-zero result or (2) the computation ${x}(x)$ does not terminate with a zero result. With either reading, you can reduce the halting problem to this one.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:07
add a comment |
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$begingroup$
You have to clarify whether ${x}(x) neq 0$ means (1) the computation ${x}(x)$ terminates and returns a non-zero result or (2) the computation ${x}(x)$ does not terminate with a zero result. With either reading, you can reduce the halting problem to this one.
$endgroup$
– Rob Arthan
Dec 6 '18 at 19:07