# Converse nonimplication

In logic, converse nonimplication 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 $P\nleftarrow Q$ .

 $P$ $Q$ $P\nleftarrow Q$ 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"

"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.

 0 1 ${\textstyle {}\sim x}$ 1 0 x
and
1 0 y 1 1 0 1 ${\textstyle y_{\vee }x}$ x
and
1 0 y 0 1 0 0 ${\textstyle y_{\wedge }x}$ x
then ${y\nleftarrow x}\!$  means
1 0 y 0 0 0 1 ${y\nleftarrow x}\!$ x
(Negation) (Inclusive or) (And) (Converse nonimplication)

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 ${rp=0}$  #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 iff ${rp=0}$ .

#### Non-commutative

• ${q\nleftarrow p=p\nleftarrow q\,}\!$  iff ${q=p\,}\!$  #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
$\mathrm {s.1}$  Definition ${q{\tilde {\leftarrow }}p=q'p\,}\!$
$\mathrm {s.2}$  Definition ${p{\tilde {\leftarrow }}q=p'q\,}\!$
$\mathrm {s.3}$  $\mathrm {s.1\ s.2}$  ${q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q'p=qp'\,}\!$
$\mathrm {s.4}$  ${q\,}\!$  ${=\,}\!$  ${q.1\,}\!$
$\mathrm {s.5}$  $\mathrm {s.4.right}$  - expand Unit element ${=\,}\!$  ${q.(p+p')\,}\!$
$\mathrm {s.6}$  $\mathrm {s.5.right}$  - evaluate expression ${=\,}\!$  ${qp+qp'\,}\!$
$\mathrm {s.7}$  $\mathrm {s.4.left=s.6.right}$  ${q=qp+qp'\,}\!$
$\mathrm {s.8}$  ${q'p=qp'\,}\!$  ${\Rightarrow \,}\!$  ${qp+qp'=qp+q'p\,}\!$
$\mathrm {s.9}$  $\mathrm {s.8}$  - regroup common factors ${\Rightarrow \,}\!$  ${q.(p+p')=(q+q').p\,}\!$
$\mathrm {s.10}$  $\mathrm {s.9}$  - join of complements equals unity ${\Rightarrow \,}\!$  ${q.1=1.p\,}\!$
$\mathrm {s.11}$  $\mathrm {s.10.right}$  - evaluate expression ${\Rightarrow \,}\!$  ${q=p\,}\!$
$\mathrm {s.12}$  $\mathrm {s.8\ s.11}$  ${q'p=qp'\ \Rightarrow \ q=p\,}\!$
$\mathrm {s.13}$  ${q=p\ \Rightarrow \ q'p=qp'\,}\!$
$\mathrm {s.14}$  $\mathrm {s.12\ s.13}$  ${q=p\ \Leftrightarrow \ q'p=qp'\,}\!$
$\mathrm {s.15}$  $\mathrm {s.3\ s.14}$  ${q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q=p\,}\!$

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

## 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.