Boolean Algebra equations can be manipulated by following a few basic rules.

Manipulation Rules

 A + B = B + A
 A * B = B * A
 (A + B) + C = A + (B + C)
 (A * B) * C = A * (B * C)
 A * (B + C) = (A * B) + (A * C)
 A + (B * C) = (A + B) * (A + C)

Equivalence Rules
 =
 A = A        (double negative)
 A + A = A
 A * A = A
     _
 A * A = 0
     _
 A + A = 1

Rules with Logical Constants
 0 + A = A
 1 + A = 1
 0 * A = 0
 1 * A = A

Many of these look identical to Matrix Operations in Linear Algebra. At any rate, this permits a circuit designer to create a circuit as it comes to their mind, then manipulate the formula to generate an equivalent circuit that does the same thing but requires less space.

This can be illustrated using the 5th manipulation rule.

Using the rule, generating an equivalent circuit that does the exact same thing, but be less complicated, can be done with reasonable ease.

In the case of CMOS, the right hand side of the formula can also be manipulated, just always remember to invert. The manipulation occurs under the invert bar.

     _________________
 D = (A * B) + (A * C)

 is the same as...

     ___________
 D = A * (B + C)

The manipulation is done the exact same way. Once there is a simplified formula, using the rules with logical constants permit the placement of values directly into the formula to see what the answer is. For example, using the above non inverted formula, C is a logical 1.

 D = A * (B + C)
 D = A * (B + 1)
 D = A * (1)            [1 + A = 1]
 D = A                  [1 * A = A]

If C is known to be a logical 1, anything OR logical 1 is always a logical 1. Since the minimum requirement is one input, once a single input is true (in this case C), the other inputs don't alter the result. On the other hand, the AND gate requires all inputs. With B+C true, the only other requirement is A. As the formula gave, D will be whatever A is.