Spreading object with optional property causes incorrect type inference when `exactOptionalPropertyTypes = false`

Personne n'a encore pris cette issue.

### 🔎 Search Terms

spread optional property undefined

### 🕗 Version & Regression Information

5.9.3 and others

### ⏯ Playground Link

https://www.typescriptlang.org/play/?ts=5.9.3#code/MYewdgzgLgBAhjAvDA3jAHgLhgFgEwwC+AsAFCiSwBGAjEqhtgK5gAmApgGYCWY7rRMhWgxgdZGgB00uABoY0ybSIwA9KphQAngAd2MEJ1F1uEGL07sATlf7wzaLDBYcefVgG4VAWk0ALOCgAcjNQG3ZgWEBQciFwCBAAG3ZJBJAAcwAKMQBKMljKGCo8bEcAfmwwJgBbKmsVCUZnNi5eO0I1DSg-UxhWEHYzMBBYHSsQViZgfTgwGGsxqxgAdz92WfZ0OEiAeR0obnA4BIAFMb0rbQAVXQHETiOIdnyRYAIGxTkFaSKVdU0bgxGV7mMwWeZ2OAORqVGrWLztXxdQIhZZjMBpeSQ0R4SToGAAfVM+KarlaAkAMuTPRLJVKZV65UhAA

### 💻 Code

```ts
const a = { x: 42 }
const b1 = { x: undefined }
const c1 = { ...a, ...b1 } // type of c1 is inferred as { x: undefined; } - that's correct ✅
console.log(c1)

const b2: { x?: number } = { x: undefined } // this does not produce an error when exactOptionalPropertyTypes=false
const c2 = { ...a, ...b2 } // type of c2 is inferred as { x: number; } - that's wrong, as c2.x _is_ undefined ❌
console.log(c2)
```

### 🙁 Actual behavior

I know that there have been several issues filed before like https://github.com/microsoft/TypeScript/issues/57086 / https://github.com/microsoft/TypeScript/issues/51755 / https://github.com/microsoft/TypeScript/issues/51253 / https://github.com/microsoft/TypeScript/issues/57408 and they all were closed referring to `exactOptionalPropertyTypes`.

However, reading the docs, the [description of `exactOptionalPropertyTypes`](https://www.typescriptlang.org/tsconfig/#exactOptionalPropertyTypes) says:

> exactOptionalPropertyTypes makes TypeScript truly enforce the definition provided as an optional property:
>
> const settings = getUserSettings();
> settings.colorThemeOverride = "dark";
> settings.colorThemeOverride = "light";
>
> // But not:
> settings.colorThemeOverride = undefined;
> > Type 'undefined' is not assignable to type '"dark" | "light"' with 'exactOptionalPropertyTypes: true'. Consider adding 'undefined' to the type of the target.

So, the option changes the way, **how `?:` is defined** from "can be not present or undefined" to "can be not present, but not set to undefined".

But even when `exactOptionalPropertyTypes=false`, I don't see a reason why TS should infer an _invalid_ type in the example given above. It _knows_ that `x` can be undefined (or not present) and it _knows_ that `exactOptionalPropertyTypes=false`, so it should really infer in this case that the prop value _can_ be undefined. For me, that would appear to be a consistent implementation of the language semantics.

Or do I miss something?

### 🙂 Expected behavior

I want to rely on TS inferring correct types, also in this case.

### Additional information about the issue

Of course, this change would be a breaking change, so there could be another option to activate it.

We have a large project and there are many places (including JSON stored in customer DBs) that would need to be adjusted if we want to set `exactOptionalPropertyTypes` to true. So that's not a solution we can quickly implement.

Commencez par l’exemple TypeScript Playground lié et comparez ses types inférés avec différents paramètres de exactOptionalPropertyTypes. Lisez les issues associées #57086, #51755, #51253 et #57408 pour comprendre les décisions précédentes. C’est terminé lorsque le résultat du spread dans le cas signalé reflète que x peut être undefined sans nécessiter l’activation de exactOptionalPropertyTypes.

Rédigé par le modèle d'indexation à partir du texte de l'issue.