Incorrect union type inference for conjunction with strictNullChecks disabled

TypeScript Version: 3.4.5

Search Terms: union type inference conjunction strictNullChecks

Code

function fn(x: number) {
  return x && 'a'
}

Expected behavior:
Return type of fn should be 0 | 'a'

Actual behavior:
Return type of fn is '' | 'a' if strictNullChecks is disabled

This is causing me an issue with this specific bit of code, where I'm checking window so my code runs both on the browser and on the client:

function fn(x: number): OrientationType | undefined {
  return window && (window.outerWidth > window.outerHeight ? 'landscape-primary' : 'portrait-primary')
}

I can replace the conjunction with a conditional expression, and I know that if strictNullChecks is disabled I should have a fallback for the return value but I'm posting here for the sake of trying to understand if this is working as intended. It also feels weird that my example compiles with strictNullChecks enabled but doesn't compile when it's disabled.

Playground Link: https://www.typescriptlang.org/play/#src=function%20fn(x%3A%20number)%3A%20OrientationType%20%7C%20undefined%20%7B%0D%0A%20%20return%20window%20%26%26%20(window.outerWidth%20%3E%20window.outerHeight%20%3F%20'landscape-primary'%20%3A%20'portrait-primary')%0D%0A%7D

Related Issues:

まず、リンクされている TypeScript Playground の例を strictNullChecks を無効にした場合と有効にした場合の両方で再現します。型チェッカーが fn(x: number) の論理積式をどのように推論するかを追跡します。無効にした設定で期待される戻り値の型 0 | 'a' が生成され、有効にした設定や window の例を壊さなければ完了です。