Incorrect error when assigning to omitted and extended record type

[WearyMonkey]

TypeScript Version:
typescript@3.2.0-dev.20181011
typescript@3.1.3
typescript@3.0.3

Search Terms:

Code

type Baa = { a: number, b: number };
type Omit<T, K> = Pick<T, Exclude<keyof T, K>>;

function foo<P extends Baa>() {
    const foo: { foo: string } & Omit<P, 'b'> = null as any;
    const props: Baa = foo;
}

Expected behavior:

No error.

Actual behavior:

test.ts:6:11 - error TS2322: Type '{ foo: string; } & Pick<P, Exclude<keyof P, "b">>' is not assignable to type 'Baa'.
  Property 'a' is missing in type '{ foo: string; } & Pick<P, Exclude<keyof P, "b">>'.

6     const props: Baa = foo;
            ~~~~~

Removing the { foo: string } & works.
Removing the Omit works.
Removing the P and using Baa directly works.

Playground Link:

https://agentcooper.github.io/typescript-play/?ts=2.9.1#code/C4TwDgpgBAQghnKBeKBvKcBcUB2BXAWwCMIAnAGiiIH5t9iyoBfAbgChRIoB5AgS2AAeACqUA0gD5kUAAp8AxgGsRlAKIAPeQBs8AEwiDFEEAHsAZlFFRJE9mzN4c84HxM4oZkycEyoEdcAQOLoAzrAIEgAUAJRobFAJUPJuIcAeXtjonibYqaR8OADmzFAAZDz8QjKUAORENVIo+FpaGGFwOCDsiUkpaWCkJmAh2PCIKNnsTGxAA

Related Issues:

This isn't just a problem with the error message, but that we don't recognize Omit<P, 'b'> as having any properties at all.

function f<T extends Record<"a" | "b", any>>(x: Record<Exclude<keyof T, 'b'>, any>): { a: any } { return x; }

T's keys are at least "a" | "b", so Exclude<keyof T, "b"> contains at least "a", so x should have a as a key. I think the type system currently does not have a notion on a lower bound on a type -- we can specify that T extends something, but not T super something. So we can't understand that keyof T will contain "a" even when "b" is excluded.