Type inference should take advice from interfaces

[na-sa-do]

Bug? Suggestion? Oh well.

TypeScript Version: 2.7.0-dev.20171024

Code

interface OneOrTwo {
    ex: 1 | 2;
}

class Example implements OneOrTwo {
    ex = 1;
}

Expected behavior: The type of Example#ex is inferred to 1 | 2 and the example compiles.

Actual behavior: The type of Example#ex is inferred to number, which is not a subtype of 1 | 2, so the example does not compile.

Suggested solution?: If a property of a class has its type inferred from a default value, and that type isn't compatible with all interfaces the class explicitly implements, try inferring the intersection of the types required by the interfaces. So:

interface OneOrTwo {
    ex: 1 | 2;
}

interface TwoOrThree {
    ex: 2 | 3;
}

class Example implements OneOrTwo, TwoOrThree {
    ex = 2; // should be inferred as type (1 | 2) & (2 | 3), i.e. 2
}

For simplicity, if that type doesn't work, the original type (number here) should be used in error messages. Maybe the message should be changed, even.

[RyanCavanaugh]

See #10570