Type narrowing of `super` when using `this`

Bug Report

🔎 Search Terms

super, this, class

🕗 Version & Regression Information

Nightly

⏯ Playground Link

https://www.typescriptlang.org/play?ts=5.2.0-dev.20230720#code/MYGwhgzhAECCB2BLAtmEAeAKgPmgbwChpjoBTMAFy2wAoKALRCALjiVQxwEp8iT+GAJwD2Ad2jxS4gKKCRgmgHJMjGABNhpCPArRUFCqUGKuAbj7EAvgWsFQkGABFhAc2pkAHofhqYCFGjuhPzkVACqtAxMrM5uETzB-CQQAK4ADkYAdKE0ZtAA9PnQKqTQFACeGdBMbAGc2NmU1HSqrP4c1FysAG7CiGoWSUPDhcX0pRVVEPTCKSBq0ABGpe2BEY3hka21HfE9fQP81tZAA

💻 Code

class Animal<T> {
    eat<T>(this: Animal<T>) {
        throw new Error('This doesnt matter');
    }
}

class Dog<T> extends Animal<T> {
    eat<U>(this: Dog<U>) {
        super.eat(); // The type is Animal<T>.eat<T>(this: Animal<T>): void
                     // The type should be Animal<U>.eat<U>(this: Animal<U>): void
    }
}

🙁 Actual behavior

super.eat() is typed as Animal<T>.eat<T>(this: Animal<T>): void

🙂 Expected behavior

super.eat() should be typed as Animal<U>.eat<U>(this: Animal<U>): void since this is declared as Dog<U> which is a subclass of Animal<U>.

One can workaround this by writing super.eat<U>(), but the signature then becomes goofy (it becomes Animal<T>.eat<U>(this: Animal<U>): void)

Start with the linked TypeScript Playground and reproduce the difference between the current and expected types for super.eat() in the generic Animal and Dog classes. Trace the compiler's handling of this types in super property access, then add a regression test and confirm that super.eat() preserves the U-specific type without requiring an explicit type argument.