Assignability rule for conditional types needs to require check types and distributivity to be identical

TypeScript Version: master (e4718564e5b2c1f2e7bbef27fc480fd8172dfdc0)

Search Terms: assignable assignability conditional type check type checkType distributive distributivity identical unsound

Code

type MyElement<A> = [A] extends [[infer E]] ? E : never;
function oops<A, B extends A>(arg: MyElement<A>): MyElement<B> { 
    return arg;  // compiles OK, expected compile error
}
oops<[number | string], [string]>(42).slice();  // runtime error

type MyAcceptor<A> = [A] extends [[infer E]] ? (arg: E) => void : never;
function oops2<A, B extends A>(arg: MyAcceptor<B>): MyAcceptor<A> { 
    return arg;  // compiles OK, expected compile error
}
oops2<[number | string], [string]>((arg) => arg.slice())(42);  // runtime error

type Dist<T> = T extends number ? number : string; 
type Aux<A extends { a: unknown }> = A["a"] extends number ? number : string;
type Nondist<T> = Aux<{a: T}>;
function oops3<T>(arg: Dist<T>): Nondist<T> {
    return arg;  // compiles OK, expected compile error
}
oops3<number | string>(42).slice();  // runtime error

Expected behavior: Compile errors as marked.

Actual behavior: Compiles OK, runtime errors as marked.

Playground Link: link

Related Issues: None found

Start with the three TypeScript examples and the linked Playground reproduction, focusing on the marked assignments that compile without errors. Investigate how conditional types compare their check types and distributivity, then add compiler tests covering the expected errors and verify that the marked runtime failures are rejected.