Union type resolving to one side of the union when spread operator is used

Bug Report

This issue may be related but they were different enough to warrant a write up here: https://github.com/microsoft/TypeScript/issues/42665

When assigning a union type to a variable, which is immediately instantiated as an object that could possibly be either type at run time, and the spread operator is used to "optionally" apply a property, the type of the variable falls to the type without the extra properties automatically.

🔎 Search Terms

Spread operator, object, union type

🕗 Version & Regression Information

This changed between versions 3.7.5 and 3.8.3.

⏯ Playground Link

Playground link with relevant code

💻 Code

interface Foo {
    bar: string;
}

type Bar = Foo & {
    baz: string;
}

const produceBoolean = () => Math.random() > 0.5;

const obj: Foo | Bar = {
    bar: 'baz',
    ...(produceBoolean() && {
        baz: 'bar'
    })
}

if ('baz' in obj) { // obj is "Foo" here
    console.log(obj); // obj is "never" here
}

🙁 Actual behavior

Instead of remaining a union type until some discrimination was made, it assumed the object was the side of the union without extra properties.

🙂 Expected behavior

It would remain possibly either or any type in the union until discriminated in the code.

Start with the linked TypeScript Playground reproduction and compare its behavior across the reported 3.7.5 and 3.8.3 versions. Trace the type-checking path for object spread into a union, then verify that the resulting value remains a union and that the 'baz' in obj check narrows it correctly.