4 ms·
I mean, the lamdbda calculus doesn't allow you to define a function!
by bodhiandpysics1 5y ago
I mean, the lamdbda calculus doesn't allow you to define a function!
- kazinator 5y agoI don't think that's true when lambda calculus is regarded as a "library" of certain objects and concepts added to mathematics. Then the assignment operator is available. You've probably seen the Y combinator given similarly to: Y = λf.(λx.f(x x))(λx.f(x x)) where this Y = business isn't part of lambda calculus per se, but it's available because this is mathematics; existing mathematics notation that doesn't conflict with lambda calculus, like being able to equate a symbol and expression, hasn't gone out the window.
- kazinator 5y agoErratum: "assignment operator" is not the best choice of words there, other than in the sense of "assign meaning" (to a hitherto unbound symbol).