Omit<A | B> works unexpectly, cannot keep union relationship

[k8w]

TypeScript Version: 3.7.x-dev.201xxxxx
4.0.3

Search Terms:
Omit Union Omit<A|B>

Code

type A = {
    type: 'A',
    valueA: string,
    common: string
}
type B = {
    type: 'B',
    valueB: string,
    common: string
}
type AB = A | B;
type NoCommon = Omit<AB, 'common'>;

let value: NoCommon = {
    type: 'A',
    valueA: 'XXX'
}

Expected behavior:
No compile error

Actual behavior:

error TS2322: Type '{ type: "A"; valueA: string; }' is not assignable to type 'Pick<AB, "type">'.
Object literal may only specify known properties, and 'valueA' does not exist in type 'Pick<AB, "type">'.
16 valueA: 'XXX'

Playground Link:

Related Issues:

[MartinJohns]

Duplicate of #39556. Search terms: omit union

Search Terms: Omit Union Omit<A|B>

A tip for the future: Something overly specific to your example, such as the types A and B, don't make good search terms. :-)

[RyanCavanaugh]

To get the behavior you wanted, the syntax is

type NoCommonHelper<T> = T extends unknown ? Omit<T, 'common'> : never;
type NoCommon = NoCommonHelper<AB>;

[typescript-bot]

This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.