conditional operator with derived and base classes with conversion operator to built-in types issue, compiler bug?

Question

This example:

struct X {};
struct Y : X {};

using CY = const Y;

true ? X() : CY(); // error

Which is explained in this answer, could be changed like this:

struct X { operator int() { return 0; } };
struct Y : X { operator int() const { return 0; } };

using CY = const Y;

true ? X() : CY(); // compiles, and ?: expression yields prvalue of int

The problem is according to the standard and linked explanation, which is based on interpretation of the standard, at clause [expr.cond] p4.3.2, which says:

otherwise, if T2 is a base class of T1, the target type is cv1 T2, where cv1 denotes the cv-qualifiers of T1;

CY() operand should be converted to type const X, which means that at clause [expr.cond] p6:

Otherwise, the result is a prvalue. If the second and third operands do not have the same type, and either has (possibly cv-qualified) class type, overload resolution is used to determine the conversions (if any) to be applied to the operands ([over.match.oper], [over.built]). If the overload resolution fails, the program is ill-formed. Otherwise, the conversions thus determined are applied, and the converted operands are used in place of the original operands for the remainder of this subclause.

already converted operand of type const X somehow calling Y::operator int() const, and it shouldn't, you can say, because of slicing. Right?

Also modified example compiles on all major compilers.