Here is the implementation of binary trees that I have seen in Martin Odersky's video lectures on Coursera, week 3:
abstract class IntSet
{
def incl(x: Int): IntSet
def contains(x: Int): Boolean
def union(other: IntSet): IntSet
}
object Empty extends IntSet
{
def contains(x: Int): Boolean = false
def incl(x: Int): IntSet = new NonEmpty(x, Empty, Empty)
def union(other: IntSet): IntSet = other
override def toString = "."
}
class NonEmpty(elem: Int, left: IntSet, right: IntSet) extends IntSet
{
def contains(x: Int): Boolean =
if (x<elem) left contains x
else if (x > elem) right contains x
else true
def incl(x: Int): IntSet =
if (x<elem) new NonEmpty(elem, left incl x, right)
else if (x > elem) new NonEmpty(elem, left, right incl x)
else this
def union(other: IntSet): IntSet =
((left union right) union other) incl elem
override def toString = "{" + left + elem + right + "}"
}
So Empty and NonEmpty follow the standards outlined by the IntSet. What I am interested in, is the union method in the NonEmpty class. I would like to understand how it works.
I have made a small depiction below to explain my thinking process:

To me it appears as if there is an infinite loop there. But I am more than certain that I have made an evaluation mistake somewhere below:
- ((L_1 U R_1) U O) incl e1
- ((L_3 U R_3) U R_1) incl e3
- (E U R_1) incl e3
- 2.
- 3.
- 2. etc. etc.