Typescript type guard for Function with Parameters

Kinda hard to explain.

type FunctionWithArguments<T> = (...args: T[]) => T;

Some in-built implementations:

type Parameters<O extends (...args: any[]) => any> = O extends (
  ...args: infer P
) => any
  ? P
  : never;

type ReturnType<O extends (...args: any[]) => any> = O extends (
  ...args: any[]
) => infer P
  ? P
  : never;