MATHEMATICS MAGAZINE Exclusive Disjunction and the Biconditional: An Even-Odd Relationship JOSEPH S . FULDA Mount Sinai School of Medicine New York, NY 10029-6574 An elementary truth table argument shows that exclusive disjunction is just the negation of the biconditional: ( P @ Q ) = l ( P o Q ) . This relationship is sometimes used to explain why inclusive, rather than exclusive, disjunction is the standard disjunction. Either disjunction can be formed from the other ( ( P V Q) = ( ( P @ Q ) @ ( P A Q ) ) ; ( P @ Q ) = ( ( P V Q ) A l ( P A Q ) ) ) , but only exclusive disjunction is the negation of another simple connective. However, while P @J Q is logically equivalent to the negation of P o Q , P @ Q @ R is logically equivalent to P o Q e R itself. (One can omit all parentheses in logical expressions involving only @J or o , since both connectives are commutative and associative.) The reason for this is that @ is a mutual exclusivity connective, whereas o is an identity connective. Hence, P @ Q @ R is true precisely when P @ Q and R have opposite truth values, which occurs precisely when P o Q and R have identical truth values. Generalizing this pattern gives strings of propositions connected by @J or o that alternate in accordance with the following identities: 1L

( A ) @ Pi = *n Pi , for n odd; i = l i = 1 We now prove these identities by mathematical induction on the number of propositions . Proof: Basis: The logical equivalence P1 @ P, = i ( P , o P,) follows directly from the truth tables for the two expressions. Induction Step: Assume the identities true for an integer n 2 2. We will show them true for n + 1. (A) n is odd. We begin with €9:=+;~~,which can be rewritten ( @:=,Pi) P,,,. By@J the basis, this is equivalent to (( @:=,Pi) o P,,+ ,). By the induction hypothesis, i this is equivalent to 1(( e f = , Pi ) e Pn+ ,). This, in turn, is just 1( f f,'pi), which concludes the induction step for case (A) and with it the proof of case (A). (B) n is even. We begin with €9r=+11~i,which can be rewritten ( €9 :=,Pi) CTJPn+ By the basis, this is equivalent to 1(( Pi) o Pn + ,). By the induction hypothesis, this is equivalent to l ( l ( e :=,pi) o P,+ ,). Since l ( 1 P o Q ) is true just when P and Q have identical truth values (i.e., l ( 1 P o Q ) = ( P o Q ) ) , this in turn yields ( f= ,Pi) o Pn+ ,, which is just a ff;pi . This concludes the induction step for case (B) and with it the proof of case (B). Acknowledgments. I would like to acknowledge the comments of an anonymous referee. I also wish to dedicate this note to Mr. Samuel Block, an outstanding teacher of mathematics, from whom I first learned to appreciate identities and their proofs.