12.15 Conditional operator

In 12.15 Conditional operator we have the rules for finding the common type of the second and third operands:

> The second and third operands, `x` and `y`, of the `?:` operator control the type of the conditional expression.
> - If `x` has type `X` and `y` has type `Y` then,
> - If `X` and `Y` are the same type, then this is the type of the conditional expression.
> - Otherwise, if an implicit conversion (§11.2) exists from `X` to `Y`, but not from `Y` to `X`, then `Y` is the type of the conditional expression.
> - Otherwise, if an implicit enumeration conversion (§11.2.4) exists from `X` to `Y`, then `Y` is the type of the conditional expression.
> - Otherwise, if an implicit enumeration conversion (§11.2.4) exists from `Y` to `X`, then `X` is the type of the conditional expression.
> - Otherwise, if an implicit conversion (§11.2) exists from `Y` to `X`, but not from `X` to `Y`, then `X` is the type of the conditional expression.
> - Otherwise, no expression type can be determined, and a compile-time error occurs.
> - If only one of `x` and `y` has a type, and both `x` and `y` are implicitly convertible to that type, then that is the type of the conditional expression.
> - Otherwise, no expression type can be determined, and a compile-time error occurs.

However the phrase "if an implicit enumeration conversion (§11.2.4) exists from `X` to `Y`" makes no sense, as an implicit enumeration conversion is a *conversion from expression*, whereas `X` is a type, not an expression. This error appears twice. I think the "from" variable should be the lower-case version, which indicates the corresponding expression.