Signatures of intersection/union types properties

Consider next example:

interface A{
    f:string;
    set_f(val:string) : void;
    get_f() : string;
}

interface B{
    f:number;
    set_f(val:number) : void;
    get_f() : number;
}

let n:number;
let str:string

let first : A&B;
first.f = n; // error, as f has type number&string. Probably OK, but probably it should be number|string, to be symmetric with set_f function.
first.f = str; // error, as f has type number&string.
n = first.f;
str = first.f;

first.set_f(n);
first.set_f(str);
n = first.get_f(); // error, typescript prefer to use first signature () => string. Shouldn't it be string&number.
str = first.get_f();

let second : A|B;
second.f = n; // Ops, no error! But it should be at least string&number here
second.f = str;// Ops, no error! But it should be at least string&number here
n = second.f; // error, typescript prefer to use first signature string. It should be string|number
str = second.f;// Ops, no error, typescript prefer to use first signature string. It should be string|number

second.set_f(n); //error, lack a call signature. It's really good, but better if it would be (string&number) => void
second.set_f(str); //error, lack a call signature. 
n = second.get_f(); // error, cool, typescript was able to correctly infer result as string|number
str = second.get_f();// error, cool, typescript was able to correctly infer result as string|number

With #9167 additional to previous was added preserving readonly flag for intersection. Should't readonly flag be reset if any of intersection member has no such flag? As intersection should extend possibilities of type, and never narrow down it.

issue にある A/B の交差と和の例を、#9167 の後で言及されている readonly のケースも含めて再現してください。まず、交差と和についてプロパティ型とメソッドシグネチャがどのように解決されるかを追跡します。代入、戻り値の型、readonly の保持に関する意図した動作について合意され、関連するチェックでカバーされれば完了です。