Typescript 创建通过删除参数来转换函数的对象文字的类型

我正在try 在类型脚本中创建一个函数，该函数接受函数的对象文字，并返回具有非常相似签名的对象文字，只有第一个参数已被删除.这里的目标是完全保留类型，因为很容易返回对象文本，其中成员的函数的参数是未知的参数array.

目标如下:

const res = doSomething('someVal', { func1: (proxy, other: string) => proxy + other });
res.func1('other string') // This knows the "other" param is a string, and returns a string.

我可以创建一个编译的结果，但输出文字的函数都是参数数组，或者我得到了某种编译器错误.以下是我的众多try 之一.

type Proxy<T> = {
  value: T;
};

/* 
 * The first parameter is always Proxy<T>.
 * The remaining parameters are what I want to keep in the return function.
 */
export type ProxyMethod<T, F extends (signalProxy: Proxy<T>, ...args: readonly any[]) => any> =
  F extends (...args: infer P) => infer R
    ? (args: P) => R
    : never;

/** This is the type of the object literal of functions that is the input */
export type ProxyMethods<T, K extends string> = {
  [Key in K]: ProxyMethod<T, (...args: unknown[]) => unknown>
};

/** Takes a ProxyMethod and creates a function without the first param. */
export type OutputMethod<F extends ProxyMethod<any, () => any>> =
  F extends ProxyMethod<infer T, infer F2>
    ? F2 extends ((signalProxy: Proxy<T>, ...args: infer P) => infer R)
      ? (...args: P) => R : never
    : never;

/** The intended output object literal */
export type OutputMethods<M extends ProxyMethods<any, string>> = { [Key in keyof M]: OutputMethod<M[Key]> };

function doSomething<T, const M extends Record<string, ProxyMethod<T, any>>>(
  value: T,
  methods: M
): OutputMethods<M> {
  /* do something */
  return methods as any; // stub, all I care about is the typings.
}

// Target signature provides too few arguments
doSomething('test', { help: (proxy: Proxy<string>, a: number) => `${a}`});

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