Exclude Falsy values from discriminated union

Question

Ideally I'd want an explicit false when there's no error and exclude all Falsy values from error when there is an error:

type Result<T, E> = { result: T; error: false } | { result?: T; error: E }

type User = {
   name: string
}

export const someFunction = (): Result<User, string> => {
   if (Math.random() > 0.5)
      return {
         error: false,
         result: {
            name: 'John',
         },
      }

   return { error: '' } //  I'd want an empty string to not be allowed
}

const someOtherFunction = () => {
   const { error, result: user } = someFunction()
   if(error === false) { //  I'd want all Falsy values to be excluded from E, so I could have a clean type guard: if(error)
        return
    }
   console.log(user) //  User | undefined, given there's no error, should be User only
}

someOtherFunction()

But this doesn't work How can I exclude all Falsy values from my right-hand side error in a way that makes the destructuring assignment work cleanly?

Here's a minimal reproducible example.