User blog comment:Pteriforever/Ultra-F notation/@comment-25337554-20151103001528/@comment-1782812-20151103063633

F + F + n expands out to F + n nested copies of F + & + n.

It's pretty easy to show that these two small cases terminate:

F + A = A ^ 2

F + (F + B) + A = (B(B + A) ^ 2 + A) ^ 2

Next, assume F + (F + (F + ... + (F + (F + B) + A) ... + A) + A) + A = X(A,B) is well-defined for any given natural numbers A and B, where there are C nested copies of F + & + A.

Then, take  F + (F + (F + ... + (F + (F + B) + A) ... + A) + A) + A with C + 1 nested copies of F + & + A. It is equal to:

F + (F + (F + ... + (F + (F^ + B) + A) ... + A) + A) + A |   F + (F + (F + ... + (F + B + A) ... + A) + A) + A

= F + (F + (F + ... + (F + (F^ + B) + A) ... + A) + A) + A | X(A,A + B)

= F + (F + (F + ... + (F + BX(A,A + B) + A) ... + A) + A)

= X(A,BX(A,A + B) + A)

Therefore, by induction, F + F + n always terminates.