Type narrowing not working as expected in else

For the following code with strictNullChecks enabled:

interface Base {
    a: {} | null;
    b: {} | null;
}

function test(base: Base): void {
    if (!base.a && !base.b) {
        const result: null = null;
    } else if (!base.a && base.b) {
        const result: {} = base.b;
    } else if (base.a && !base.b) {
        const result: {} = base.a;
    } else {
        const result: {} = base.b;
    }
}

I’m seeing this error on the last result:

Type '{} | null' is not assignable to type '{}'.
  Type 'null' is not assignable to type '{}'.

I expect it to work as the else implies:

   !(!base.a && !base.b) && !(!base.a && base.b) && !(base.a && !base.b) 
 = (base.a || base.b) && (base.a || !base.b) && (!base.a || base.b)
 = (base.a || base.b) && (base.a) && (base.b)
      note: !base.b and !base.a are canceled out as the expression cannot be true if one
            of them is false
 = base.a && base.b

TS play link

[DanielRosenwasser]

Sorry, I don’t know why this was labeled as a question.

[Tyriar]

@DanielRosenwasser I think this is labelled incorrectly and shouldn’t have been closed, it never really got an answer on why or why not from a maintainer.