Converse nonimplication

Converse nonimplication in a general Boolean algebra is defined as q ↚ p = q ′ p {\textstyle 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 ∼ {\textstyle \sim } as complement operator, ∨ {\textstyle \vee } as join operator and ∧ {\textstyle \wedge } as meet operator, build the Boolean algebra of propositional logic.

∼ x {\textstyle {}\sim x} 1 0 x 0 1	and	y 1 1 1 0 0 1 y ∨ x {\textstyle y_{\vee }x} 0 1 x	and	y 1 0 1 0 0 0 y ∧ x {\textstyle y_{\wedge }x} 0 1 x	then y ↚ x {\displaystyle \scriptstyle {y\nleftarrow x}\!} means	y 1 0 0 0 0 1 y ↚ x {\displaystyle \scriptstyle {y\nleftarrow x}\!} 0 1 x
(Negation)		(Inclusive or)		(And)		(Converse nonimplication)

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 c {\displaystyle \scriptstyle {^{c}}\!} (co-divisor of 6) as complement operator, ∨ {\displaystyle \scriptstyle {_{\vee }}\!} (least common multiple) as join operator and ∧ {\displaystyle \scriptstyle {_{\wedge }}\!} (greatest common divisor) as meet operator, build a Boolean algebra.

x c {\displaystyle \scriptstyle {x^{c}}\!} 6 3 2 1 x 1 2 3 6	and	y 6 6 6 6 6 3 3 6 3 6 2 2 2 6 6 1 1 2 3 6 y ∨ x {\displaystyle \scriptstyle {y_{\vee }x}\!} 1 2 3 6 x	and	y 6 1 2 3 6 3 1 1 3 3 2 1 2 1 2 1 1 1 1 1 y ∧ x {\displaystyle \scriptstyle {y_{\wedge }x}} 1 2 3 6 x	then y ↚ x {\displaystyle \scriptstyle {y\nleftarrow x}\!} means	y 6 1 1 1 1 3 1 2 1 2 2 1 1 3 3 1 1 2 3 6 y ↚ x {\displaystyle \scriptstyle {y\nleftarrow x}\!} 1 2 3 6 x
(Co-divisor 6)		(Least common multiple)		(Greatest common divisor)		(x's greatest divisor coprime with y)

r ↚ ( q ↚ p ) = ( r ↚ q ) ↚ p {\displaystyle r\nleftarrow (q\nleftarrow p)=(r\nleftarrow q)\nleftarrow p} if and only if r p = 0 {\displaystyle rp=0} #s5 (In a two-element Boolean algebra the latter condition is reduced to r = 0 {\displaystyle r=0} or p = 0 {\displaystyle p=0} ). Hence in a nontrivial Boolean algebra converse nonimplication is nonassociative. ( r ↚ q ) ↚ p = r ′ q ↚ p (by definition) = ( r ′ q ) ′ p (by definition) = ( r + q ′ ) p (De Morgan's laws) = ( r + r ′ q ′ ) p (Absorption law) = r p + r ′ q ′ p = r p + r ′ ( q ↚ p ) (by definition) = r p + r ↚ ( q ↚ p ) (by definition) {\displaystyle {\begin{aligned}(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{aligned}}}

Clearly, it is associative if and only if r p = 0 {\displaystyle rp=0} .

• q ↚ p = p ↚ q {\displaystyle q\nleftarrow p=p\nleftarrow q} if and only if q = p {\displaystyle q=p} #s6. Hence converse nonimplication is noncommutative.

• 0 is a left neutral element ( 0 ↚ p = p {\displaystyle 0\nleftarrow p=p} ) and a right absorbing element ( p ↚ 0 = 0 {\displaystyle {p\nleftarrow 0=0}} ).
• 1 ↚ p = 0 {\displaystyle 1\nleftarrow p=0} , p ↚ 1 = p ′ {\displaystyle p\nleftarrow 1=p'} , and p ↚ p = 0 {\displaystyle p\nleftarrow p=0} .
• Implication q → p {\displaystyle q\rightarrow p} is the dual of converse nonimplication q ↚ p {\displaystyle q\nleftarrow p} #s7.