Understanding TypeScript template literal type with `${any}${any}`

Question

Can someone explain why this:

type NotEmpty<T extends string> = T extends `${any}${any}` ? T : never;

declare function something<T extends string>(x: NotEmpty<T>): void;

something("hello")
something(""); // Argument of type 'string' is not assignable to parameter of type 'never'.(2345)

achieves the obvious goal of the type and yet this does not:

type NotEmpty<T extends string> = T extends `${string}${string}` ? T : never;

declare function something<T extends string>(x: NotEmpty<T>): void;

something("hello")
something(""); // okay

and also this does not:

type NotEmpty<T extends string> = T extends `${any}` ? T : never;

declare function something<T extends string>(x: NotEmpty<T>): void;

something("hello")
something(""); // okay

Answer 1

I'm surprised the first one works to be honest: Intuitively, you'd think '' extends string extends any.
¯\(?)/¯

Here's a working type that in case any future person needs one that is straightforward.

type NonEmpty<T extends string> = '' extends T ? never : T;

But this feels like cheating because it explicitly uses empty string and doesn't answer your question.

However, If you expand out either the ${any}${any} or ${string}${string} using infer, they both yield your expected result.

type NonEmpty<T extends string> = T extends `${infer First}${infer Rest}` ?
 First extends string ? Rest extends string ? T : never : never : never;

which only adds to my confusion. I know there are some simplifications to types when it comes to template literals. I think the best path to an answer would be to open an issue on the Typescript repo and hope that Anders (the Typescript guy) or someone on the team clarifies.