Converse nonimplication in a general
Boolean algebra is defined as q \nleftarrow p=q'p. Example of a 2-element Boolean algebra: the 2 elements {0,1} with 0 as zero and 1 as unity element, operators \sim as complement operator, \vee as join operator and \wedge as meet operator, build the Boolean algebra of
propositional logic. Example of a 4-element Boolean algebra: the 4 divisors {1,2,3,6} of 6 with 1 as zero and 6 as unity element, operators \scriptstyle{ ^{c}}\! (co-divisor of 6) as complement operator, \scriptstyle{_\vee}\! (least common multiple) as join operator and \scriptstyle{_\wedge}\! (greatest common divisor) as meet operator, build a Boolean algebra.
Properties Non-associative r \nleftarrow (q \nleftarrow p) = (r \nleftarrow q) \nleftarrow p if and only if rp = 0
#s5 (In a
two-element Boolean algebra the latter condition is reduced to r = 0 or p=0). Hence in a nontrivial Boolean algebra converse nonimplication is
nonassociative. \begin{align} (r \nleftarrow q) \nleftarrow p &= r'q \nleftarrow p & \text{(by definition)} \\ &= (r'q)'p & \text{(by definition)} \\ &= (r + q')p & \text{(De Morgan's laws)} \\ &= (r + r'q')p & \text{(Absorption law)} \\ &= rp + r'q'p \\ &= rp + r'(q \nleftarrow p) & \text{(by definition)} \\ &= rp + r \nleftarrow (q \nleftarrow p) & \text{(by definition)} \\ \end{align} Clearly, it is associative if and only if rp=0.
Non-commutative • q \nleftarrow p=p \nleftarrow q if and only if q = p
#s6. Hence converse nonimplication is
noncommutative.
Neutral and absorbing elements • is a left
neutral element (0 \nleftarrow p=p) and a right
absorbing element ({p \nleftarrow 0=0}). • 1 \nleftarrow p=0, p \nleftarrow 1=p', and p \nleftarrow p=0. • Implication q \rightarrow p is the dual of converse nonimplication q \nleftarrow p
#s7. ==Computer science==