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Hi

Could a function in FOL take functions as arguments? FOL only limits on the order of the individuals being quantified, but if an expression doesn't does not involve quantifying over second-order or higher terms, would it still be valid in FOL? Say,

f(g)$f(g)$ where

f::('a=>'b)=>'c $f : (A \to B) \to C$ and g::'a=>'b$g : A \to B$.

So another way to put my question is that does the rule of formation in FOL say that f(g) $f(g)$ is valid as long as g $g$ is in the domain of f, $f$, despite the type of f? i.e. f could $f$? Could $f$ be a 4th order function and would still be valid in FOL?

Thanks
show/hide this revision's text 2 added 239 characters in body

Hi

Could a function in FOL take functions as arguments? FOL only limits on the order of the individuals being quantified, but if an expression doesn't involve quantifying over second-order or higher terms, would it still be valid in FOL? Say,

f(g)

where

f::('a=>'b)=>'c and g::'a=>'b

So another way to put my question is that does the rule of formation in FOL say that f(g) is valid as long as g is in the domain of f, despite the type of f? i.e. f could be a 4th order function and would still be valid in FOL?

Thanks
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Second-order term in first-order logic?

Hi

Could a function in FOL take functions as arguments? FOL only limits on the order of the individuals being quantified, but if an expression doesn't involve quantifying over second-order or higher terms, would it still be valid in FOL? Say,

f(g)

where

f::('a=>'b)=>'c and g::'a=>'b