Converse nonimplication

In logic, converse nonimplication[1] is a logical connective which is the negation of converse implication (equivalently, the negation of the converse of implication).

Venn diagram of

P\nleftarrow Q

(the red area is true)

Definition

Converse nonimplication is notated $P\nleftarrow Q$ , or $P\not \subset Q$ , and is logically equivalent to $\neg (P\leftarrow Q)$

Truth table

The truth table of $P\nleftarrow Q$ .[2]

$P$	$Q$	$P\nleftarrow Q$
T	T	F
T	F	F
F	T	T
F	F	F

Notation

Converse nonimplication is notated ${\textstyle p\nleftarrow q}$ , which is the left arrow from converse implication ( ${\textstyle \leftarrow }$ ), negated with a stroke ( $/$ ).

Alternatives include

${\textstyle p\not \subset q}$ , which combines converse implication's $\subset$ , negated with a stroke ( $/$ ).
${\textstyle p{\tilde {\leftarrow }}q}$ , which combines converse implication's left arrow ( ${\textstyle \leftarrow }$ ) with negation's tilde ( ${\textstyle \sim }$ ).
Mpq, in Bocheński notation

Properties

falsehood-preserving: The interpretation under which all variables are assigned a truth value of 'false' produces a truth value of 'false' as a result of converse nonimplication

Natural language

Grammatical

"p from q."

Classic passive aggressive: "yeah, no"

Rhetorical

"not A but B"

Boolean algebra

Converse Nonimplication in a general Boolean algebra is defined as ${\textstyle q\nleftarrow p=q'p}$ .

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.

${\textstyle {}\sim x}$	$1$	$0$
$x$	$0$	$1$

and

$y$
$1$	$1$	$1$
$0$	$0$	$1$
${\textstyle y_{\vee }x}$	$0$	$1$	$x$

and

$y$
$1$	$0$	$1$
$0$	$0$	$0$
${\textstyle y_{\wedge }x}$	$0$	$1$	$x$

then

\scriptstyle {y\nleftarrow x}\!

means

$y$
$1$	$0$	$0$
$0$	$0$	$1$
$\scriptstyle {y\nleftarrow x}\!$	$0$	$1$	$x$

(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 $\scriptstyle {^{c}}\!$ (codivisor 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.

$\scriptstyle {x^{c}}\!$	$6$	$3$	$2$	$1$
$x$	$1$	$2$	$3$	$6$

and

$y$
$6$	$6$	$6$	$6$	$6$
$3$	$3$	$6$	$3$	$6$
$2$	$2$	$2$	$6$	$6$
$1$	$1$	$2$	$3$	$6$
$\scriptstyle {y_{\vee }x}\!$	$1$	$2$	$3$	$6$	$x$

and

$y$
$6$	$1$	$2$	$3$	$6$
$3$	$1$	$1$	$3$	$3$
$2$	$1$	$2$	$1$	$2$
$1$	$1$	$1$	$1$	$1$
$\scriptstyle {y_{\wedge }x}$	$1$	$2$	$3$	$6$	$x$

then

\scriptstyle {y\nleftarrow x}\!

means

$y$
$6$	$1$	$1$	$1$	$1$
$3$	$1$	$2$	$1$	$2$
$2$	$1$	$1$	$3$	$3$
$1$	$1$	$2$	$3$	$6$
$\scriptstyle {y\nleftarrow x}\!$	$1$	$2$	$3$	$6$	$x$

(Codivisor 6)

(Least common multiple)

(Greatest common divisor)

(x's greatest divisor coprime with y)

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{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 $rp=0$ .

Non-commutative

$q\nleftarrow p=p\nleftarrow q$ if and only if $q=p$ #s6. Hence Converse Nonimplication is noncommutative.

Neutral and absorbing elements

$0$ 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.

Converse Nonimplication is noncommutative
Step	Make use of	Resulting in
s.1	Definition	$\scriptstyle {q{\tilde {\leftarrow }}p=q'p\,}\!$
s.2	Definition	$\scriptstyle {p{\tilde {\leftarrow }}q=p'q\,}\!$
s.3	s.1 s.2	$\scriptstyle {q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q'p=qp'\,}\!$
s.4		$\scriptstyle {q\,}\!$	$\scriptstyle {=\,}\!$	$\scriptstyle {q.1\,}\!$
s.5	s.4.right - expand Unit element		$\scriptstyle {=\,}\!$	$\scriptstyle {q.(p+p')\,}\!$
s.6	s.5.right - evaluate expression		$\scriptstyle {=\,}\!$	$\scriptstyle {qp+qp'\,}\!$
s.7	s.4.left = s.6.right	$\scriptstyle {q=qp+qp'\,}\!$
s.8		$\scriptstyle {q'p=qp'\,}\!$	$\scriptstyle {\Rightarrow \,}\!$	$\scriptstyle {qp+qp'=qp+q'p\,}\!$
s.9	s.8 - regroup common factors		$\scriptstyle {\Rightarrow \,}\!$	$\scriptstyle {q.(p+p')=(q+q').p\,}\!$
s.10	s.9 - join of complements equals unity		$\scriptstyle {\Rightarrow \,}\!$	$\scriptstyle {q.1=1.p\,}\!$
s.11	s.10.right - evaluate expression		$\scriptstyle {\Rightarrow \,}\!$	$\scriptstyle {q=p\,}\!$
s.12	s.8 s.11	$\scriptstyle {q'p=qp'\ \Rightarrow \ q=p\,}\!$
s.13		$\scriptstyle {q=p\ \Rightarrow \ q'p=qp'\,}\!$
s.14	s.12 s.13	$\scriptstyle {q=p\ \Leftrightarrow \ q'p=qp'\,}\!$
s.15	s.3 s.14	$\scriptstyle {q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q=p\,}\!$

Implication is the dual of Converse Nonimplication
Step	Make use of	Resulting in
s.1	Definition	$\scriptstyle {\operatorname {dual} (q{\tilde {\leftarrow }}p)\,}\!$	$\scriptstyle {=\,}\!$	$\scriptstyle {\operatorname {dual} (q'p)\,}\!$
s.2	s.1.right - .'s dual is +		$\scriptstyle {=\,}\!$	$\scriptstyle {q'+p\,}\!$
s.3	s.2.right - Involution complement		$\scriptstyle {=\,}\!$	$\scriptstyle {(q'+p)''\,}\!$
s.4	s.3.right - De Morgan's laws applied once		$\scriptstyle {=\,}\!$	$\scriptstyle {(qp')'\,}\!$
s.5	s.4.right - Commutative law		$\scriptstyle {=\,}\!$	$\scriptstyle {(p'q)'\,}\!$
s.6	s.5.right		$\scriptstyle {=\,}\!$	$\scriptstyle {(p{\tilde {\leftarrow }}q)'\,}\!$
s.7	s.6.right		$\scriptstyle {=\,}\!$	$\scriptstyle {p\leftarrow q\,}\!$
s.8	s.7.right		$\scriptstyle {=\,}\!$	$\scriptstyle {q\rightarrow p\,}\!$
s.9	s.1.left = s.8.right	$\scriptstyle {\operatorname {dual} (q{\tilde {\leftarrow }}p)=q\rightarrow p\,}\!$

Computer science

An example for converse nonimplication in computer science can be found when performing a right outer join on a set of tables from a database, if records not matching the join-condition from the "left" table are being excluded.[3]

References

Lehtonen, Eero, and Poikonen, J.H.
Knuth 2011, p. 49
http://www.codinghorror.com/blog/2007/10/a-visual-explanation-of-sql-joins.html

Knuth, Donald E. (2011). The Art of Computer Programming, Volume 4A: Combinatorial Algorithms, Part 1 (1st ed.). Addison-Wesley Professional. ISBN 978-0-201-03804-0.

External links

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[1] Lehtonen, Eero, and Poikonen, J.H.

[Knuth-2] Knuth 2011, p. 49

[3] ttp://www.codinghorror.com/blog/2007/10/a-visual-explanation-of-sql-joins.html

Logical connectives
Tautology/True $\top$
Alternative denial (NAND gate) $\uparrow$ Converse implication $\leftarrow$ Implication (IMPLY gate) $\rightarrow$ Disjunction (OR gate) $\lor$
Negation (NOT gate) $\neg$ Exclusive or (XOR gate) $\not \leftrightarrow$ Biconditional (XNOR gate) $\leftrightarrow$ Statement (Digital buffer)
Joint denial (NOR gate) $\downarrow$ Nonimplication (NIMPLY gate) $\nrightarrow$ Converse nonimplication $\nleftarrow$ Conjunction (AND gate) $\land$
Contradiction/False $\bot$
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