The way we usually prove that something is unique, i.e. that no other object has the property we are talking about, is to show that if two objects have the property, then they are equal. In this case, we will let x and z be arbitrary but fixed and show that if x and z are both additive identities, then x = z.

1: ∀y[x + y = y] ∧ ∀y[z + y = y]   [Assumption A - i.e. assuming x and z both are "additive identitites"]

2: ∀y[x + y = y]                   _________________________________________________________________

3: ∀y[z + y = y]                   _________________________________________________________________

4: x + z = z                       _________________________________________________________________

5: z + x = x                       _________________________________________________________________

6: z + x = x + z                   _________________________________________________________________
 
7: x = z                           _________________________________________________________________

8: ∀y[x + y = y] ∧ ∀y[z + y = y] => x = z __________________________________________________________

Since x and z are arbitrary but fixed, we deduce that for all x and z, if x and z are both additive identities, then x = z. Thus the addititive identity is unique.

0:                             [let x, y and y' be arbitrary but fixed]

1: x + y = 0 ∧ x + y' = 0      [Assumption]


2: y + 0 = y + 0               


3: y + (x + y) = y + (x + y')  


4: (x + y) + y = (x + y) + y'  


5: 0 + y = 0 + y'              


6: y = y'                      


7: x + y = 0 ∧ x + y' = 0 => y = y'                


8: all x,y,y'[x + y = 0 ∧ x + y' = 0 => y = y']  [arbitrary but fixed arguments become forall's]