Allow use of `infer` in the type parameter of a type declaration

Suggestion

For many type declarations, it shouldn't be necessary to use a conditional type which repeats a constraint already found in type parameter.

For example, instead of:

type ConstructorParameters<T extends abstract new (...args: any    ) => any> =
                           T extends abstract new (...args: infer P) => any ? P : never;

This would be better expressed as:

type ConstructorParameters<T extends abstract new (...args: infer P) => any> = P;

🔍 Search Terms

infer type parameter conditional type type declaration

✅ Viability Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, new syntax sugar for JS, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

⭐ Suggestion

See above

📃 Motivating Example

TypeScript's own built-in types have other examples where constraints are repeated in both a type parameter and conditional type. I've also seen cases where developers just don't want to repeat constraints, so they define:

type RecordValueType<T> =
    T extends Record<any, infer V> ? V : never;

And then incorrect usage results in a type of never, rather than an error.

💻 Use Cases

• Improved readability
• Less repetition when authoring type declarations
• Less frustration when consuming lazily-declared types

Start with the motivating examples and the proposed infer syntax in the issue, then review how TypeScript currently handles type parameter constraints and conditional types. Done means type declarations can infer a type parameter directly from its constraint without repeating a conditional type, with the stated readability and validation behavior preserved.