Type Inference In Switch Statement Not Work With &&

TypeScript Version: 4.0.0-dev.20200512

Search Terms: switch, inference, switch and inference

Code

interface I1 {
    type: 'I1';
    value: string;
}

interface I2 {
    type: 'I2';
    value: number;
}

type T = I1 | I2;

let t: T | undefined;

if (t && t.type == 'I1') {
    t.value; // Type string
}

switch (t?.type) {
    case 'I1':
        t.value; // Type string
        break;
}

switch (t && t.type) {
    case 'I1':
        t.value; // Uh oh, error
        t!.value; // Uh oh, string | number
        break;
}

Expected behavior:
The compile should understand that t && t.type is like as t?.type except that has also other falsy values.

Actual behavior:
The compiler does not infer from such expression.

Playground Link: https://www.typescriptlang.org/play/?ts=4.0.0-dev.20200512#code/JYOwLgpgTgZghgYwgAgJIEZkG8BQz-JgCeADhAFzIDkGVA3HgQG5wA2ArhcgM5hSgBzBgF8cOUJFiIUqAEzZG+YmUo1Z9RchYcuIdgFsARtBFjlKACrIAvGkwAfNLIY5WEMIUpXH7EABMIGFAIPxdgGGQACg8AMhjCADpzG1sadCoASgUCRO1OOmQAekLkC1IUXn4QARxRHG4Ad2AwBAALKLAAfiTyrNwchDhuFDSqck0csAS8iALi0vKePkEJgkMoCDgAa1N6ppb26OQ4xPM+zUHh6lpxnMnptnyikoBVdoB7VoAaZGgod6gqyUAEIHjo5q8Pt8llUBMhHHojNAgch1psdrUcEA

Start with the linked TypeScript Playground reproduction and compare the narrowing behavior of the three shown conditions. Trace the compiler's control-flow analysis for switch expressions using t && t.type; done means the final branch narrows t consistently with the expected behavior.