Type refinement doesn't work on destructured properties

[StevenLangbroek]

TypeScript Version: 3.7 (tried nightly too)

Search Terms:

• refinement destructuring
• refinement destructure
• refinement destructured property
• refinement union destructured

Code

type Result = {
    __typename: 'AuthError'
} | {
    __typename: 'OtherError'
} | {
    __typename: 'User';
    firstName: string;
    lastName: string;
}

const getResult = (): Promise<Result> => fetch('some/url').then(res => res.json());

getResult().then(result => {
    const { __typename } = result;

    if (__typename === 'AuthError') {
        return;
    }

    if (__typename === 'OtherError') {
        return;
    }

    // result isn't refined into `__typename: User` here:
    console.log(result.firstName);
})

Expected behavior:

I'd expected this to be refined regardless of whether testing the destructured discriminator or accessing it through the argument.

Actual behavior:

When destructuring the type is not refined.

Playground Link:

Playground

Related Issues:

[ajafff]

searching for "refinement destructuring" brought up issue #35283 which is basically the same.
TypeScript cannot narrow a variable based on another variable's refinement.

[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.