How should empty arrays and union types work together in 1.6?

[osdm]

var p = new Array<string>();
var q = true ? p : [];
q[0]

My reasoning follows, please correct me if I'm wrong.
According to 1.5 spec, type of q[0] was calculated in the following way:

1. Take union type of Array<string> and empty array (Array<Undefined>).
2. Reduce that union type, which gives us Array<string>, because string is a supertype of undefined.
3. Widen that type (changes nothing).
4. Type of q[0] is string.

According to 1.6 spec, type of q[0] should be calculated in the following way:

1. Take union type of Array<string> and empty array (Array<Undefined>).
2. No reduce is happening.
3. Widen that type, resulting in (Array<string> | Array<any>).
4. Type of q[0] is (string | any).

But that would break "no implicit any". Where did I make a mistake?

[DanielRosenwasser]

@ahejlsberg, can you check into this?

[ahejlsberg]

@osdm You're right, we no longer do subtype reduction in union types, however we still do deduplication as discussed in #4115. This has yet to make it into the spec. In the example above, Array<Undefined> is considered a duplicate of Array<string> so the type of the ?: operation simply becomes Array<string>.