Transitive nature of equals method

Question

The contract for equals(object) method specifies 4 properties to follow: Reflexive, Symmetric, Transitive and Consistent. While I understand the danger of not f

Answer 1

Assume three objects a,b,c with

a == a, b == b, c == c (reflexive)
a == b, b == a
b == c, c == b
a != c, c != a

^{(Pseudocode, x == y stands for x.equals(y)).}

Now, let's add the objects to a set:

Set s = new HashSet(); // Set implementation doesn't matter
s.add(b); // s = [b]
s.add(a); // s doesn't change, because a == b
s.add(c); // s doesn't change, because c == b

In contrast, if we were to add them in a different order:

Set s = new HashSet();
s.add(a); // s = [a]
s.add(b); // s doesn't change, because b == a
s.add(c); // s = [a,c], because c != a

That is clearly counter-intuitive and doesn't match the behavior one would expect of a set. For example, it would mean that the union of two sets (i.e. the state of s after s.addAll(someOtherSet)) could depend on the implementation (order of elements) of someOtherSet.