Type inference - code after early return from an 'if' should infer the same type as the 'else' block would #8886

Version 1.8

First of all I'm sorry if I get the jargon wrong, I'm new to typescript (I tried to search for this issue and failed, I may have used the wrong search terms).

When using type guards, code in an else block uses the condition in the if to infer the type of the variable. However if the if contains an early return the rest of the function should be treated as the else of the if from we we returned.

Code

// A self-contained demonstration of the problem follows...
type Foo = { a: string };
function isFoo(x): x is Foo {
    return x && typeof(x.a) === 'string';
}

type Bar = { b: string };

function buz1(arg: Foo|Bar) : string {
    if (isFoo(arg))
        return arg.a;  // OK arg inferred to be Foo
    else
        return arg.b; // OK arg inferred to be Bar
}

function buz2(arg: Foo|Bar) : string {
    if (isFoo(arg)) 
        return arg.a;  // OK arg inferred to be Foo

    return arg.b;  // ** Error b isn't a member of Foo|Bar **
}

Expected behavior:
function buz2 should compile with no error.
Actual behavior:
Error when accessing member b of arg even though arg can be inferred to be of type Bar (as it is in buz1.else).