Type inference/narrowing lost after assignment

TypeScript Version: 3.1

Search Terms: type inference, type guard, narrowing, lost, assignment

Code

let a: unknown = 'x';

if (typeof a === 'string') {
  // a inferred as `string`
  a = a.substr(0, 5);  // (method) String.substr(from: number, length?: number): string
  // a inferred as `unknown`
  a.length;            // Failure: Object is of type 'unknown'.
}

Expected behavior: This should compile without an error.

Actual behavior: Line 7 fails with: Object is of type 'unknown'.

Playground Link: https://www.typescriptlang.org/play/index.html#src=let%20a%3A%20unknown%20%3D%20'x'%3B%0D%0A%0D%0Aif%20(typeof%20a%20%3D%3D%3D%20'string')%20%7B%0D%0A%20%20%2F%2F%20a%20inferred%20as%20%60string%60%0D%0A%20%20a%20%3D%20a.substr(0%2C%205)%3B%0D%0A%20%20%2F%2F%20a%20inferred%20as%20%60unknown%60%0D%0A%20%20a.length%3B%0D%0A%7D%0D%0A

Related Issues: #18840, #19955, #26673

TypeScript Playground で最小限の例を再現し、関連する issue #18840、#19955、#26673 と動作を比較してください。typeof ガード内の代入によって string への絞り込みが維持され、最終的な a.length 式がエラーなしでコンパイルされれば、作業は完了です。