To install click the Add extension button. That's it.

The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.

4,5
Kelly Slayton
Congratulations on this excellent venture… what a great idea!
Alexander Grigorievskiy
I use WIKI 2 every day and almost forgot how the original Wikipedia looks like.
Live Statistics
English Articles
Improved in 24 Hours
Languages
Recent
Show all languages
What we do. Every page goes through several hundred of perfecting techniques; in live mode. Quite the same Wikipedia. Just better.
.
Leo
Newton
Brights
Milds

# Converse nonimplication

Venn diagram of ${\displaystyle P\nleftarrow Q}$
(the red area is true)

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

## Definition

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

### Truth table

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

 ${\displaystyle P}$ ${\displaystyle Q}$ ${\displaystyle P\nleftarrow Q}$ T T F T F F F T T F F F

## Notation

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

Alternatives include

• ${\displaystyle \textstyle {p\not \subset q}}$, which combines converse implication's ${\displaystyle \subset }$, negated with a stroke (/).
• ${\displaystyle \textstyle {p{\tilde {\leftarrow }}q}}$, which combines converse implication's left arrow(${\displaystyle \textstyle {\leftarrow }}$) with negation's tilde(${\displaystyle \textstyle {\sim }}$).
• 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 ${\displaystyle \scriptstyle {y\nleftarrow x}\!}$ means
1 0 y 0 0 0 1 ${\displaystyle \scriptstyle {y\nleftarrow x}\!}$ 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 ${\displaystyle \scriptstyle {^{c}}\!}$ (codivisor 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.

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

### Properties

#### Non-associative

${\displaystyle \scriptstyle {r\nleftarrow (q\nleftarrow p)=(r\nleftarrow q)\nleftarrow p}}$ iff ${\displaystyle \scriptstyle {rp=0}}$ #s5 (In a two-element Boolean algebra the latter condition is reduced to ${\displaystyle \scriptstyle {r=0}}$ or ${\displaystyle \scriptstyle {p=0}}$). Hence in a nontrivial Boolean algebra Converse Nonimplication is nonassociative.

{\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 iff ${\displaystyle \scriptstyle {rp=0}}$.

#### Non-commutative

• ${\displaystyle \scriptstyle {q\nleftarrow p=p\nleftarrow q\,}\!}$ iff ${\displaystyle \scriptstyle {q=p\,}\!}$ #s6. Hence Converse Nonimplication is noncommutative.

#### Neutral and absorbing elements

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

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

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

## 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

• 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.
Basis of this page is in Wikipedia. Text is available under the CC BY-SA 3.0 Unported License. Non-text media are available under their specified licenses. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc. WIKI 2 is an independent company and has no affiliation with Wikimedia Foundation.