Infering type of optional parameter of lambda function is broken when passed to generic function.

TypeScript Version: 3.7.0-beta, nightly and older ones

Search Terms:
optional parameter infer lambda generic
Code

declare function noop<T>(val: T): T;

const f = (a: string, b = 23) => b;

// reports "Parameter 'b' implicitly has an 'any' type."
// broken since v3.5.1, older versions work well
const foo = noop((a: string, b = 23) => b);  

// (a: string, b?: number) => number
const bar = noop(f);
 
declare function noop2<T extends (...params: any[]) => any>(val: T): T;

// (a: string, b?: any) => any (BUG)
const foo2 = noop2((a = 23) => a);  // broken on all versions

// (a: string, b?: number) => number
const bar2 = noop2(f);

Expected behavior:

foo* and bar* should all have the same type.

Actual behavior:

In the first case, the compiler complains that it can't infer the type of optional argument of lambda forwarded to noop, and in the second case it silently assigns any type to b instead of number

Playground Link:

https://www.typescriptlang.org/play/?ts=Nightly&ssl=11&ssc=23&pln=1&pc=1#code/CYUwxgNghgTiAEAzArgOzAFwJYHtX1RxwAcAeAFQD4AKANyggC55yBKZ8gbgFgAoPlOmx4k1KMwDOGGFlQBzADTwARvAC88AEwBmVvADe8OBmQx8UTvAC+fPmDxSkRdQSLFqYydNmKVLnXpqlPBQrJZ2DhgqsC6EJNSIYXzwfKCQsAiCmLj4ccSaFPAgAB4YIKjAEvDUAHR1xLBQALYSzFCoAJ4A2gC6gcHtHTT0TCzsLDz8vPaojohEmrFumh7i8FIy8kqqGgHqA2Ep05HRMIsaeSuJnEA

Notes
The mentioned versions of typescript are not exact, I just used TypeScript playground and the set of available versions there.

まず、リンク先の TypeScript Playground の例、特にオプションのラムダパラメーターを指定した noop と noop2 の呼び出しを再現します。これらの式に対するコンパイラーのジェネリック推論と文脈的型付けを追跡します。foo、bar、foo2、bar2 が implicit any エラーなしで同等のパラメーター型と戻り値型を推論できれば完了です。