Property is not assignable to the same property in generic base type

まだ誰も着手していません。

### 🔎 Search Terms

Property is not assignable to the same property in generic base type

### 🕗 Version & Regression Information

-This is the behavior in every version I tried, and I reviewed

### ⏯ Playground Link

https://www.typescriptlang.org/zh/play/?ts=5.8.3#code/IYIwzgLgTsDGEAJYBthjAgSgCwPYIG8AoBBUSGeBABwEsATAc1oDswAeAMQH0BRBAKYAPCAJb0MLAK4BbEAKgAaBDwBSAPgAUIKFLABHKQMgAuFd1UIAPud4BKM2uu2A3EQC+RFGgwAhZFIAZoEKgiJiElh4hCQ0DMxsXHxhouKSsvJK5hraugZGptnOPPaOFsXJscSkpFACEFJQLKQ6eobGEG6knp5AA

### 💻 Code

```ts
abstract class Rho {
abstract pidgins(brusquest: F_J | F_E): F_J | F_E;
}
class Bluffer extends Rho {
pidgins(brusquest: F_J | F_E): F_J | F_E
{
return brusquest;
}
}

```

### 🙁 Actual behavior

Property 'pidgins' in type 'Bluffer' is not assignable to the same property in base type 'Rho'.
Type '(brusquest: F_E | F_J) => F_E | F_J' is not assignable to type '(brusquest: F_E | F_J) => F_E | F_J'. Two different types with this name exist, but they are unrelated.
Type 'number | F_E | F_J' is not assignable to type 'F_E | F_J'.
Type 'number' is not assignable to type 'F_E | F_J'.(2416)

### 🙂 Expected behavior

No error

### Additional information about the issue

The following works with no error:
```
abstract class Rho {
abstract pidgins(brusquest: F_J | F_E): F_J | F_E;
abstract pidgins(brusquest: F_J | F_E): F_J | F_E;
}
class Bluffer extends Rho {
pidgins(brusquest: F_J | F_E): F_J | F_E
{
return brusquest;
}
}

```

リンクされた TypeScript Playground の例を再現することから始め、Rho のジェネリックメソッドと Bluffer の実装を比較します。ジェネリックメソッドの互換性に関する型チェックの経路を追跡します。再現結果でエラーが報告されず、関連する代入可能性の動作を壊していなければ完了です。

索引モデルが issue の本文から書いたものです。