kniga 7 - Probability and Statistics 1 - Sheynin, Oscar

It is important to indicate that all the rules concerning the combination of propositions(having to do with the symbol and) are necessary conclusions from the rules about thejoining of the propositions (involving the symbol or). And very remarkable is the duality thatexists here: The rules regarding the symbols “or” and “and” are absolutely identical so thatall the formulas persist when the symbols are interchanged if{only}the impossible and thetrue propositions, O and , at the same time also replace each other.Indeed, suffice it to observe all the above to note that the sole difference between the twosets of rules is that (A and ) = A, (A and O) = O, whereas (A or ) = and (A or O) = A.1.1.8. The Principle (Axiom 1.6) of UniquenessTo complete our system, I introduce one more principle (the principle of uniqueness as Ishall call it) underpinning the notion of negation. It can be stated thus: If proposition iscompatible with all the propositions of a totality (excepting O), it is true: = .Definition of negation: The join A of all propositions incompatible with A is called thenegation of A.Corollary 1.14. Ω = O.Corollary 1.15. O = (.Corollary 1.16. If x = O, then x = . Indeed, all the propositions (excepting O) arecompatible with x and, consequently, on the strength of the principle of uniqueness, x = .Corollary 1.17. If x = then x = O. Indeed, since is the join of the propositionsincompatible with x, it itself possesses the same property because of the restrictive principleand therefore x = O.Any set of propositions whose join is is called solely possible.Theorem 1.5. Propositions A and A are solely possible and incompatible; that is, (A orA ) = , (A and A ) = O.Indeed, any proposition is either a particular case of A or compatible with A; therefore,because of the principle of uniqueness 1 , (A or A ) = . On the other hand, since A is the joinof propositions incompatible with A, (A and A ) = O.Theorem 1.6. A negation of proposition A is equivalent with A: neg( A ) = A.It is sufficient to prove that A = A 1 always follows from(A or B) = (A 1 or B), (A and B) = (A 1 and B).Indeed, however,A 1 = [A 1 and (B or A 1 )] = [A 1 and (B or A)] =[(A 1 and B) or ( A 1 and A)] = [(A and B) or (A 1 and A)] =[A and (B or A 1 )] = [A and (B or A)] = A.Definition. If (A or B) = B,then the join C of all the propositions which are particular cases of B and incompatible withA is called the complement of A to B. Thus, C = (B and A ). Conversely, A is a complementof C to B. Indeed, (A or C) = B, (A and C) = O. Suppose that A 1 is a complement of C to B,then we would also have (A 1 or C) = B, (A 1 and C) = O and A = A 1 . Therefore, A = (B andC ).Corollary 1.18. Neg (A and B) = ( A or B ). Indeed,{[(A and B) or A ] or B } = [( and B) or B ] = ,