如何在TypeScript函数式管道中通过高阶转换器自动推断类型？

需求背景 #

我正在构建一个模拟器，使用fp-ts的pipe让对象通过多个函数逐步转换。为提升代码表现力，我采用高阶函数创建转换器，期望代码能写成如下形式：

pipe(
    { name: "Nick" }, 
    merge({ age: 34 }), 
    merge({ job: "programmer" }), 
    merge({ married: true })
);

理想状态下，通过为pipe指定类型实现类型安全，示例如下：

type Guy = { name: string };
type GuyWithAge = Guy & { age: number };
type GuyWithAgeAndJob = GuyWithAge & { job: string };
type GuyWithAgeAndJobAndMarried = GuyWithAgeAndJob & { married: boolean };

pipe<Guy, GuyWithAge, GuyWithAgeAndJob, GuyWithAgeAndJobAndMarried>(
    { name: "Nick" },
    merge({ age: 34 }), 
    merge({ job: "programmer" }), 
    merge({ married: true })
);

// 最终结果: { name: "Nick", age: 34, job: "programmer", married: true }

我希望TypeScript能自动推断每一步的类型，仅接受类型间的差异，确保对象转换的类型安全。

当前实现与问题 #

我已通过泛型和TypeScript工具类型实现核心功能，但存在冗余问题：每次调用merge时都需要显式指定泛型类型。代码如下：

// Merge 类似 A & B，但会用B的属性覆盖A的同名属性，等同于 {...A, ...B}
type Merge<First extends object, Second extends object> = Omit<
  First,
  keyof Second
> &
  Second;

// Delta 是与 From 合并后得到 To 的对象类型
export type Delta<From extends object, To extends object> = {
  [K in keyof To as K extends keyof From
    ? To[K] extends From[K]
      ? never
      : K
    : K]: To[K];
};

const merge =
  <First extends object, Second extends object>(
    delta: Delta<First, Second>,
  ) =>
  (obj: First): Merge<First, Delta<First, Second>> => {
    const merged = {
      ...obj,
      ...delta,
    };
    return merged;
  };

type Guy = { name: string };
type GuyWithAge = Guy & { age: number };
type GuyWithAgeAndJob = GuyWithAge & { job: string };
type GuyWithAgeAndJobAndMarried = GuyWithAgeAndJob & { married: boolean };

const guy = pipe<Guy, GuyWithAge, GuyWithAgeAndJob, GuyWithAgeAndJobAndMarried>(
  { name: "Nick" },
  merge<Guy, GuyWithAge>({ age: 34 }),
  merge<GuyWithAge, GuyWithAgeAndJob>({ job: "programmer" }),
  merge<GuyWithAgeAndJob, GuyWithAgeAndJobAndMarried>({ married: true })
);

我的问题是：有没有办法让TypeScript自动推断pipe中每个merge调用的正确泛型，从而无需手动指定类型？

补充分析 #

我进一步拆解了问题：pipe期望接收A => B类型的函数。当传入merge<First, Second>(delta)时，它返回(obj: First) => Merge<First, Delta<First, Second>>。

也就是说，我向pipe提供的函数类型是(obj: First) => Merge<First, Delta<First, Second>>，而pipe需要的是A => B，理论上A应映射到First，B应映射到Merge<First, Delta<First, Second>>（理想情况下可简化为Second，但不确定该简化是否正确）。

我想理解为何TypeScript无法自动推断A对应First、B对应Second，以及如何修复代码以实现预期的自动推断。

内容的提问来源于stack exchange，提问作者Nick Manning