Typescript Omit seems to transform an union into an intersection

Question

I have this type

type Cartesian = { kind: 'cartesian'; x: number; y: number; }
type Polar = { kind: 'polar'; angle: number; distance: number }
type Movement = { name: string } & (Cartesian | Polar);

that I can use like this

const move = (movement: Movement) => { /* whatever */ };
move({ name: 'top right', kind: 'cartesian', x: 10, y: 10 });
move({ name: 'left', kind: 'polar', angle: Math.PI, distance: 10 });

but for some reason, I can't use it like this

const unnamedMove = (unnamedMovement: Omit<Movement, 'name'>) => {
    move({ name: 'default', ...unnamedMovement })
}

because TS throws a 2345:

Argument of type '{ kind: "cartesian" | "polar"; name: string; }' is not assignable to parameter of type 'Movement'.
  Type '{ kind: "cartesian" | "polar"; name: string; }' is not assignable to type '{ name: string; } & Polar'.
    Type '{ kind: "cartesian" | "polar"; name: string; }' is missing the following properties from type 'Polar': angle, distance

I don't understand this.

If I am not mistaken Omit<Movement, 'name'> should be equivalent to the union type Cartesian | Polar, which would make { name: 'default', ...unnamedMovement } a Movement, and all things should work.

However, it looks like TS infers Omit<Movement, 'name'> as if it was the union type Cartesian & Polar, thus the error.

Is it a bug or a mistake from my part?

playground link

Answer 1

You probably want Distributive conditional types. From the docs

When conditional types act on a generic type, they become distributive when given a union type.

This means that you can declare something like this

type DOmit<T, K extends string> = T extends unknown
  ? Omit<T, K>
  : never;

and then use it like this

const unnamedMove = (unnamedMovement: DOmit<Movement, "name">) => {
    move({ name: 'default', ...unnamedMovement })
}

Playground here