`typeof ... === "undefined"` check on discriminated union of `undefined` and object type doesn't narrow correctly

[fmmoret]

Bug Report

🔎 Search Terms

typeof narrow, typeof discriminated union

🕗 Version & Regression Information

Affects all playground versions.

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• This is the behavior in every version I tried, and I reviewed the FAQ for entries mentioning "typeof" or "narrow". Closest is this https://github.com/Microsoft/TypeScript/wiki/FAQ#type-guards

The FAQ entry is TODO, but I don't think it applies here, the union is not & 'd with the type guard type at all.

⏯ Playground Link

Playground link with relevant code

💻 Code

https://www.typescriptlang.org/play?ts=4.8.0-beta#code/MYewdgzgLgBAhjAvDA3gX3hVBTATrkXALlThOlwEswBzNAGhl2wgFcAbKE1sAE2wBm1bLwwAfVDDwFiPfkLAjGzNpxIoARiTCsAthrxo0AKHbZYWmBWo0A3KfMxgRHfrz3KAgBRQAngAdsEAF4ADppQhgAQkRkAHI5QWFeOIBKVGMYLJgNJDCI3FC4WxgAelKYMDh8EAB3EWMMbHYIbAzspzy4UJUOKFCNEvLKkFgqmvreRuNPL26C6NiYRIURdJRM7NzkeZrC4rKK8YJJxqkWto2O4C6elj6BoaPqk4a0IA

const a = {} as {error: {a: string}, result: undefined} | { error:undefined, result: {b: number}}
let b: string;
let c:number;
if(typeof a.error !== 'undefined') {
    b = a.error.a; // narrowed
} else {
    c = a.result.b; // not narrowed
}
if(a.error !== undefined) {
    b = a.error.a; // narrowed
} else {
    c = a.result.b; // narrowed
}

A similar example that I also expected to work but actually fails for both cases:
https://www.typescriptlang.org/play?ts=4.8.0-beta#code/MYewdgzgLgBAhjAvDA3gX3hVBTATrkXALlThOlwEswBzNAGhl2wgFcAbKEsVgWwCM8MDAB9UMPAWIweAvI2ZtOJFP3JQqtNGgBQ7bLDUwK1GgG49BmMCKzBuC-tiUIAOT72kMABQAPEqxgANZgIADuYACUJL4wLjIeQogAfDBQAJ4ADtggAGYwsYhFMADkdnglFpS53hnZefAAdJKEMACExWWJuCWRqDowgzD8XnDN+ISNcGYwAPSzMnATYdgAJjoY2OwQ2P1D1qONihxQjfwz8zIgsGBLBCvrutXebS7ucrjeYy24kX0oAyGI2Q3wmuCmFwWt2Waw2Em2uwB+2Ah2OnDOkKuNzu4VhaCAA

const a = {} as {error: {a: string}, result: number } | { error: number, result: {b: string}}
let b: string;
let c:number;
let isNumber = (x: unknown): x is number => typeof x === 'number';
if(typeof a.error !== 'number') {
    b = a.error.a; // narrowed
} else {
    c = a.result.b; // not narrowed
}
if(!isNumber(a.error)) {
    b = a.error.a; // narrowed
} else {
    c = a.result.b; // not narrowed
}

🙁 Actual behavior

typeof based narrowing only narrowed one code flow branch.

🙂 Expected behavior

typeof based narrowing should narrow both sides of the branch.

[RyanCavanaugh]

Shorter repro:

declare const a: {error: {a: string}, result: undefined} | { error:undefined, result: {b: number}}
if (a.error === undefined) {
    a.result.b; // OK
}
if (typeof a.error === 'undefined') {
    a.result.b; // Not OK, but should be
}