Type guard on conditional parameter type

Question

Why is this code not working ?

Typescript could easily guess that a variable of type (T extends '1' ? '1' : never) will never be false, so that NonFalse<TypeWithCondition<T>> is exactly the same than true | (T extends '1' ? '1' : never).

Is it a Typescript bug ? Should it be posted as a Feature Request ?

type TypeWithCondition<T> = boolean | (T extends '1' ? '1' : never);

type NonFalse<T> = Exclude<T, false>;

const logNonFalse = <T>(b: NonFalse<TypeWithCondition<T>>) => {
    console.log(b);
};

const test1 = <T>(a: TypeWithCondition<T>) => {
    if (a === false) throw new Error("Can't be false");
    logNonFalse(a);
    // Error : Argument of type 'true | ([T] extends ["1"] ? "1" : never)' is not assignable to parameter of type 'true'.
    //  Type '[T] extends ["1"] ? "1" : never' is not assignable to type 'true'.
    //    Type '"1"' is not assignable to type 'true'.(2345)
};

Answer 1

Typescript is not yet able to handle that. Cf https://github.com/microsoft/TypeScript/issues/38863