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I prefer it when Typescript tries to provide static type checking for Javascript rather than providing checking for how they wish Javascript was. `(x: number, y: string) => x == y` is an error because 'this condition will always return false'. In what language? There are linter rules for enforcing === over ==; Typescript having incorrect type definitions to try and make sure you only use == like === seems like a really poor solution to that.

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Aren't you implicitly assuming that TS's number === JS's Number and TS's string === JS's String? You could say == is extended for TS's types(number/string), but the JS == still exists: x as any == y as any;

I can see what you're saying, but I think that argument would be a lot stronger if == compiled to ===. If you want an operator with ==='s semantics, it's there; providing == and claiming it behaves like === just seems like an opportunity for annoying surprises if something slips through the cracks. Since == doesn't compile to ===, I wouldn't use it unless I understood its semantics and wanted them.

I'm guessing you can't use `(x: number, y: string) => x === parseInt(y)`?

As long as you feed parseInt it's second radix parameter you can :p

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