`as const` breaks intra expression inference

还没有人认领这个 Issue。

### 🔎 Search Terms

const context expression assertion inference type parameter

### 🕗 Version & Regression Information

- This is the behavior in every version I tried

### ⏯ Playground Link

https://www.typescriptlang.org/play/?ts=6.0.0-dev.20250806#code/CYUwxgNghgTiAEAzArgOzAFwJYHtXwxAGcMAeAFQBp4ARAPgApYBzALngG8AoeeLVAA7IM7GgG4e8ATBzBkYEOyYw2tAJTwAvHXjkJvMHiLIAtovjLV5DdvipTAIxAwJAXzWiJXQ6hLwHMgDWIPiaBMQYDNy8-EIi8ACMAAxJlJLSsvLmljY6LAB0GDgAyhgw-MwMamkGRqbZLLnwBRAhzBgAFmmuzUTwPiRqXgMY8ADuODCBFVrhJFGSscLsyanpMnIKSo1aeSqFJWUVVTX9dWbbKk0tbZ3dQ1xAA

### 💻 Code

```ts
declare function test(arg: {
input: D;
produce: (arg: D) => T;
consume: (arg: T) => number;
}): D;

const broken = test({
input: 100,
produce: (arg) => arg.toString(), // 'arg' is of type 'unknown'.(18046)
consume: (arg) => arg.length,
} as const);

const working = test({
input: 100,
produce: (arg) => arg.toString(), // `arg` inferred as `number`
consume: (arg) => arg.length,
});
```

### 🙁 Actual behavior

`as const` is the only difference between those two and its presence breaks assigned parameter type in `produce`

### 🙂 Expected behavior

I think `as const` should not impact inference this way and both should be inferred just OK. It's just that the `broken` case should infer `100` instead of `number` for the `D` type parameter

### Additional information about the issue

_No response_

从链接的 TypeScript Playground 开始，比较 `broken` 和 `working` 示例，然后跟踪 `as const` 断言如何影响 `produce` 的上下文类型和推断。完成的标准是：在使用断言的情况下，回调参数能够在没有 `unknown` 的情况下完成推断，并保留对 `D` 预期的字面量 `100` 推断。

由索引模型根据 Issue 内容生成。