





















ab + cd = (ab) + (cd) , a 
b + c = (a  b) + c , abc
= ((ab)c) .
Examples: The first three types of substitution, appled to a + b = b + a , give e g
The fourth type of substitution, applied to the expression (a
 d)(b + c) , gives e g the formula
(a  d)(b + c) = (a  d)(c
+ b) , since b + c = c + b
is a formula.
As an example of the fifth type, replace (a  2b)
in (a  2b)(c + d) by A
, giving A(c + d) .
Then the formula A(c + d) = Ac +
Ad gives, on backsubstitution, the formula
(a  2b)(c + d) = (a  2b)c
+ (a  2b)d .
Another method: x = (ax)/a = b/a
. However, the first method can be use more generally in rings of a certain
kind (domains) provided b is a multiple of a. Then the ninth
axiom does not hold and is not needed.
We will use the nonexistence of zerodivisors to solve the equation
x² = 1 . Of course, i and i are solutions
but we must show that these are the only ones. The equation implies