Does using epsilon in comparison of floating-point break strict-weak-ordering?

Does following class breaks strict-weak-ordering (in comparison to regular std::less (So ignoring edge case values such as Nan))

struct LessWithEpsilon
{
    static constexpr double epsilon = some_value;
    bool operator() (double lhs, double rhs) const
    {
        return lhs + epsilon < rhs;
    }
};

LessWithEpsilon lessEps{};

From https://en.wikipedia.org/wiki/Weak_ordering#Strict_weak_orderings

4. Transitivity of incomparability: For all x,y,z in S, if x is incomparable with y (meaning that neither x < y nor y < x is true) and if y is incomparable with z, then x is incomparable with z.

Similarly, from https://en.cppreference.com/w/cpp/named_req/Compare

If equiv(a, b) == true and equiv(b, c) == true, then equiv(a, c) == true

With {x, y, z} = {0, epsilon, 2 * epsilon}, that rule is broken:

• !lessEps(x, y) && !lessEps(y, x) && !lessEps(y, z) && !lessEps(z, y) but lessEps(x, z).
• equiv(x, y) == true and equiv(y, z) == true but equiv(x, z) == false (as x + epsilon < z)

So, that class breaks strict-weak-ordering.