# Converse nonimplication

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

## Definition

$_{p\not\subset q}\!$ which is the same as $_{\sim(p\subset q)}\!$

### Truth table

The truth table of $_{p\not\subset q}\!$.[2]

p q $_{\not\subset}\!$
T T F
T F F
F T T
F F F

### Venn diagram

The Venn Diagram of "It is not the case that B implies A" (the red area is true)

## 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

## Symbol

Alternatives for $_{p\not\subset q}\!$ are

• $_{p\tilde{\leftarrow}q}\!$: $_{\tilde{\leftarrow}}\!$ combines Converse implication's left arrow($_{\leftarrow}\!$) with Negation's tilde($_{\sim}\!$).
• $_{Mpq}\!$: uses prefixed capital letter.
• $_{p\nleftarrow q}\!$: $_{\nleftarrow }\!$ combines Converse implication's left arrow($_{\leftarrow}\!$) denied by means of a stroke($_{/}\!$).

"not A but B"

## Boolean algebra

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.

 $_{0}\!$ $_{1}\!$ $_{\sim x}\!$ $_{1}\!$ $_{0}\!$ $_{x}\!$
and
$_{1}\!$ $_{0}\!$ $_{y}\!$ $_{1}\!$ $_{1}\!$ $_{0}\!$ $_{1}\!$ $_{y_\vee x}\!$ $_{x}\!$
and
$_{1}\!$ $_{0}\!$ $_{y}\!$ $_{0}\!$ $_{1}\!$ $_{0}\!$ $_{0}\!$ $_{y_\wedge x}\!$ $_{x}\!$
then $_{y \nleftarrow x}\!$ means
$_{1}\!$ $_{0}\!$ $_{y}\!$ $_{0}\!$ $_{0}\!$ $_{0}\!$ $_{1}\!$ $_{y \nleftarrow x}\!$ $_{x}\!$
(Negation) (Inclusive Or) (And) (Converse Nonimplication)

[4] 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 $_{ ^{c}}\!$ (codivisor of 6) as complement operator, $_{_\vee}\!$ (least common multiple) as join operator and $_{_\wedge}\!$ (greatest common divisor) as meet operator, build a Boolean algebra.

 $_{1}\!$ $_{2}\!$ $_{3}\!$ $_{6}\!$ $_{x^c}\!$ $_{6}\!$ $_{3}\!$ $_{2}\!$ $_{1}\!$ $_{x}\!$
and
$_{6}\!$ $_{3}\!$ $_{2}\!$ $_{1}\!$ $_{y}\!$ $_{6}\!$ $_{6}\!$ $_{6}\!$ $_{6}\!$ $_{3}\!$ $_{6}\!$ $_{3}\!$ $_{6}\!$ $_{2}\!$ $_{2}\!$ $_{6}\!$ $_{6}\!$ $_{1}\!$ $_{2}\!$ $_{3}\!$ $_{6}\!$ $_{y_\vee x}\!$ $_{x}\!$
and
$_{6}\!$ $_{3}\!$ $_{2}\!$ $_{1}\!$ $_{y}\!$ $_{1}\!$ $_{2}\!$ $_{3}\!$ $_{6}\!$ $_{1}\!$ $_{1}\!$ $_{3}\!$ $_{3}\!$ $_{1}\!$ $_{2}\!$ $_{1}\!$ $_{2}\!$ $_{1}\!$ $_{1}\!$ $_{1}\!$ $_{1}\!$ $_{y_\wedge x}\!$ $_{x}\!$
then $_{y \nleftarrow x}\!$ means
$_{6}\!$ $_{3}\!$ $_{2}\!$ $_{1}\!$ $_{y}\!$ $_{1}\!$ $_{1}\!$ $_{1}\!$ $_{1}\!$ $_{1}\!$ $_{2}\!$ $_{1}\!$ $_{2}\!$ $_{1}\!$ $_{1}\!$ $_{3}\!$ $_{3}\!$ $_{1}\!$ $_{2}\!$ $_{3}\!$ $_{6}\!$ $_{y \nleftarrow x}\!$ $_{x}\!$
(Codivisor 6) (Least Common Multiple) (Greatest Common Divisor) (x's greatest Divisor coprime with y)

### Properties

#### Non-associative

$_{r \nleftarrow (q \nleftarrow p)=(r \nleftarrow q) \nleftarrow p}\!$ iff $_{rp=0}\!$ [5] (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 \qquad \qquad \qquad ~~~~ \text{(by definition)} \\ &= (r'q)'p \qquad \qquad \qquad ~~~~~~ \text{(by definition)} \\ &= (r + q')p \qquad \qquad ~~~~~~~~~ \text{(De Morgan's laws)} \\ &= (r + r'q')p \qquad \qquad ~~~~~~~ \text{(Absorption law)} \\ &= rp + r'q'p \\ &= rp + r'(q \nleftarrow p) \qquad ~~~~~~~~ \text{(by definition)} \\ &= rp + r \nleftarrow (q \nleftarrow p) \qquad ~~~~ \text{(by definition)} \\ \end{align}

Clearly, it is associative iff $_{rp=0}\!$.

#### Non-commutative

• $_{q \nleftarrow p=p \nleftarrow q\,}\!$ iff $_{q=p\,}\!$ [6]. 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 $_{q \nleftarrow p}\!$ [7].

[6]

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

[7]

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

## 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]