Typescript 被推断为原始类型的交集的扩充类型的交集

我正在try 从外部依赖项扩充接口. 原因是外部d.ts文件没有指定完整的返回类型. 在下面的示例中，doStuff()实际上存在于由Bar.giveList()和Baz.giveList()返回的对象上，但原始声明不包含此信息. 以下是使用declare global的问题的重新构建(我实际上是在try 增加命名模块，但当在单个文件中使用两个declare global块时，该问题似乎类似).

export type MyList<T> = T[] & {
    doStuff(): void
}

declare global { // Original declaration from external dependency
    export interface Bar {
        giveList(): string[]
    }
    export interface Baz {
        giveList(): string[]
    }
}

declare global { // My attempt of refining some return types
    export interface Bar {
        giveList(): MyList<string>
    }
    export interface Baz {
        giveList(): MyList<string>
    }
}

let bar        = {} as Bar
let baz        = {} as Baz
let barbaz     = {} as (Bar | Baz)

let barList    = bar.giveList()     // infered type: MyList<string> --- good
let bazList    = baz.giveList()     // infered type: MyList<string> --- good
let barbazList = barbaz.giveList()  // infered type: string[]       --- why not MyList<string>?

barList.doStuff()                   // good
barList[0]                          // good
bazList.doStuff()                   // good
bazList[0]                          // good
barbazList.doStuff()                // ERROR: Property 'doStuff' does not exist on type 'string[]'.(2339)
barbazList[0]                       // good

既然Bar.giveList()和Baz.giveList()返回MyList<string>，为什么(Bar|Baz).giveList()不也返回MyList<string>呢？