# Really confused about the chord algorithm

I'm looking at the pseudocode for chord algorithm here:

// create a new Chord ring.
n.create()
predecessor := nil
successor := n

// join a Chord ring containing node n'.
n.join(n')
predecessor := nil
successor := n'.find_successor(n)

// called periodically. n asks the successor
// about its predecessor, verifies if n's immediate
// successor is consistent, and tells the successor about n
n.stabilize()
x = successor.predecessor
if x ∈ (n, successor) then
successor := x
successor.notify(n)

// n' thinks it might be our predecessor.
n.notify(n')
if predecessor is nil or n'∈(predecessor, n) then
predecessor := n'

// called periodically. refreshes finger table entries.
// next stores the index of the finger to fix
n.fix_fingers()
next := next + 1
if next > m then
next := 1
finger[next] := find_successor(n+{\displaystyle 2^{next-1}}2^{next-1});

// called periodically. checks whether predecessor has failed.
n.check_predecessor()
if predecessor has failed then
predecessor := nil


I'm struggling to make sense of it. My question specifically is: when exactly will the predecessor field EVER be non-null?

Consider a case where a node id 1 joins. Its successor is 1, the predecessor is nil.

Now say node with 10 joins. Its successor is 1, predecessor is nil.

The stabilization routine depends on the predecessor being non-nil, which in this case it isn't both nodes.

So how does the algorithm proceed at this point?