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Has $n$ parameters?

i.e.

0-valued function: $f(\emptyset)=2$

1-valued function: $f(x)=x$

2-valued function: $f(x,y)=x+y$

3-valued function: $f(x,y,z)=x+y+z$

Not sure

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    $\begingroup$ Pretty sure that "0/1-valued" means that the output is always 0 or 1. But context is everything, so please let us know where you found this. $\endgroup$ – Morgan Rodgers Mar 18 at 0:35
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    $\begingroup$ Just the phrase "0/1-valued function" suggests that the function's codomain is the set $\{0,1\}$. $\endgroup$ – vadim123 Mar 18 at 0:36
  • $\begingroup$ If I hear "n-valued function" I think the mean $f(x)$ may have multiple outputs (violating the standard definition of "function" which is practically writ in stone that a well-defined function has a single output for each point in the domain). For example $f:[-1,1]\to [0,2\pi)$ via $f(x) = \theta$ if $\sin \theta x$ is a 2-value "function". But in this context I really actually have no idea. $\endgroup$ – fleablood Mar 18 at 0:37
  • $\begingroup$ "retty sure that "0/1-valued" means that the output is always 0 or 1." and "Just the phrase "0/1-valued function" suggests that the function's codomain is the set {0,1}". D'oh! That is almost certainly the case.... $\endgroup$ – fleablood Mar 18 at 0:37
  • $\begingroup$ It's from a class on Theory of Computation, topic: Recursive and Recursively Enumerable Sets. "total" here means there's always an output (never undefined) so I think 1st commenter is right. $\endgroup$ – A_for_ Abacus Mar 18 at 0:40
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$\phi_x:{\Bbb N}_0^k\rightarrow {\Bbb N}_0$ is a total 0/1-valued function means that the domain of $\phi_x$ is ${\Bbb N}_0^k$ and the range is $\phi({\Bbb N}_0^k)\subseteq \{0,1\}$.

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