When you have an union type and you want to infer one of the type, the common way is using type guard

interface Bird {
    fly();
    layEggs();
}

interface Fish {
    swim();
    layEggs();
}

function getSmallPet(pet: Fish | Bird) {
  // use a type guard function
  if (isFish(pet)) {
    // inside this block, pet is Fish type
    return pet.swim(); 
  }

  return pet.fly();
}

// create a type guard function
function isFish(pet: Fish | Bird): pet is Fish {
  return (pet as Fish).swim !== undefined;
}

But since Typescript 2.7, we could use in operator which give us simpler code without need to create a function.

function getSmallPet(pet: Fish | Bird) {
  if ("swim" in pet) {
    return pet.swim(); // pet: Fish
  }

  ...
}

Easy peasy!