Theorem 2. [(p => ).(q => )) => (p => I) is a tautology. To show that Theorem 2 is a tautology begin by applying the definition and get the formula [(p = q)^(q=P)]=>(p=Y)=-(((p =>g)^(q=>r)]1 - (p =>r)) (1.1) Apply the definition for each instance of the implication operator, > 50 (p =9) -- (PA) (9=>r)=-(-) (par)-(A-) Now, replace the right hand side of statement 1.1. [(p=q) ( => '))=(p=>r)--((-(A-2)^(-(1-1)-(-(p^--))) (1.2) Homework assignment is to complete the truth table for this tautology. I will post a message on the e- group with the details