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:

Empieza reproduciendo los ejemplos del TypeScript Playground enlazado con strictNullChecks desactivado y activado. Rastrea cómo el comprobador de tipos infiere la expresión de conjunción en fn(x: number). Se considera terminado cuando la configuración desactivada produce el tipo de retorno esperado 0 | 'a' sin romper la configuración activada ni el ejemplo de window.