union of pattern template literals intersected with optional brand mutually assignable to one with incompatible brand

まだ誰も着手していません。

### 🔎 Search Terms

brand, flavor, pattern template literal, placeholder, union, intersection, optional, assignable

### 🕗 Version & Regression Information

- This changed between versions 4.3.0-dev.20210330 and 4.3.0-dev.20210331
- This changed in commit or PR #43440, I think

### ⏯ Playground Link

[Playground link](https://www.typescriptlang.org/play/?ts=6.0.3#code/C4TwDgpgBAKgtmAMgS2AVQHbIPYagXigAMAGAEgG8BnYAJ2QwHMBfIqAH2IEZKb6nWAWABQoSFABCtAIYYAJgA0CsBCnRZcUAGRQKUaQH4AXFABEAD1NRmAbhFjoU2XICay+ElSYceHXsMmpiBWtiJyEADGADbStNARuDRQAEYy8i4mTul2wgkYSanOCplpisqF2VAA9FVQADwAtA1QGNhQELS02LQivcIY0nAQVGDSEdAActjqProiUFAOKp7AyqS8dAwsRDkLS1llhB5q2rr6xmaW1ruL4I6lbkeqqKf+F0EhN+HRsfGJqxVXCVnC4bnkCqVipJIeUHjZqrVGs0Ol0esIRMw+gMhiMxtAAMrYIbAAAWWwAolEqNAKPNbuJjl4NBgAPK0QnEslMSnUtbkaibARsThEHgC-jbDhmKhEiCkilUiCmOn7GFPFbeXBsjlyrmMHnQPznQJXULCPZ3aEg9zPGZa9my+XcxWvY1mYLXOnfGJxKDggEPYHZOn+lKQoOHMMg+E1epNdqdbp9ZhAA)

### 💻 Code

```ts
type TmpLitUnion = `0${string}` | `1${string}`
type BrandX = TmpLitUnion & { a?: "x" };
type BrandY = TmpLitUnion & { a?: "y" };
declare const brandY: BrandY;
const brandX: BrandX = brandY; // <-- no error
```

### 🙁 Actual behavior

`BrandX` and `BrandY` are mutually assignable, even though they have an incompatible optional `a` property.

### 🙂 Expected behavior

There should be an error when trying to assign a `BrandY` to a `BrandX` or vice versa.

### Additional information about the issue

This is from [a Stack Overflow question](https://stackoverflow.com/q/79929227/2887218). Looks like optional brands (this is called... "flavored" I guess?) don't distinguish between unions-of-pattern-template-literals. If you get rid of the union, you get the expected error:

```ts
type TmpLit = `0${string}`;
type BrandX = TmpLit & { a?: "x" };
type BrandY = TmpLit & { a?: "y" };
declare const brandY: BrandY;
const brandX: BrandX = brandY; // <-- error
```

Or if you add something else to the union, you get the expected error:
```ts
type TmpLitUnionOrSomethingElse = `0${string}` | `1${string}` | "somethingElse"
type BrandX = TmpLitUnionOrSomethingElse & { a?: "x" };
type BrandY = TmpLitUnionOrSomethingElse & { a?: "y" };
declare const brandY: BrandY;
const brandX: BrandX = brandY; // <-- error
```

Or if the brand is not optional, etc.

Not sure if this is a bug or a design limitation, or somehow intentional.

リンクされた TypeScript Playground で課題を再現し、その後、issue にある union なしおよび optional brand なしの例と比較してください。optional properties と交差する pattern template literals の union に対する代入可能性を調査してください。完了条件は、BrandX と BrandY が相互に代入可能でなくなり、報告されている比較ケースが引き続き正しいことです。

索引モデルが issue の本文から書いたものです。