1. this is called the DeMorgans theorems:

 A + B = A  B and A  B = A + B Prove DeMorgans theorems with a truth table of the form 

2. Prove that the two equations for E in the example starting on

page B-7 are equivalent by using DeMorgans theorems and the axioms shown on page B-7. 

Here is the example showen

Logic Equations Show the logic equations for the logic functions, D, , and F, described in the previous example. EXAMPLE Heres the equation for D: ANSWER D = A + B + C F is equally simple: F = A-B-C E is a little tricky Think of it in two parts: what must be true for E to be true (two of the three inputs must be true), and what cannot be true (all three cannot be true). Thus we can write E as E = ((A  B) + (A  C) + (B - C))  (A-B-C) We can also derive E by realizing that E is true only if exactly two of the inputs are true. Then we can write E as an OR of the three possible terms that have two true inputs and one false input: E = (A- B-C) + (A- C-B) + (B-C- A) Proving that these two expressions are equivalent is explored in the exercises. In Verilog, we describe combinational logic whenever possible using the assign statement, which is described beginning on page B-23. We can write a definition for E using the Verilog exclusive- OR operator as assign E = (A A B A C) * (A + B + C) * ( A * B * C ) , which is yet another way to describe this function. D and F have even simpler representations, which are just like the corresponding C code: D = A I B I C and F = A & B & C.