3 ms·
And applying the lambda to itself results in loops, afair. Edit: It was the Omega operator that did that.
by Akronymus 5y ago
And applying the lambda to itself results in loops, afair.
Edit: It was the Omega operator that did that.
- kmill 5y agoomega = Y id where id = λx. x (though Y Y does not normalize either, infinite looping like omega)