# 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).

## Definition

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

### Truth table

The truth table of $A\nleftarrow B$.
| $A$ | $B$ | $A\nleftarrow B$ |
| F | F | F |
| F | T | T |
| T | F | F |
| T | T | F |

## Notation

Converse nonimplication is notated p↚q{\textstyle p\nleftarrow q}, which is the left arrow from converse implication (←{\textstyle \leftarrow }), negated with a stroke (/).
Alternatives include
* p⊄q{\textstyle p\not \subset q}, which combines converse implication's $\subset$, negated with a stroke (/).
* p←$^~$q{\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

Example,
If it rains (P) then I get wet (Q), just because I am wet (Q) does not mean it is raining, in reality I went to a pool party with the co-ed staff, in my clothes (~P) and that is why I am facilitating this lecture in this state (Q).

### Rhetorical

Q does not imply P.

### Colloquial
## Boolean algebra

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 $\scriptstyle {y\nleftarrow x}\!$ 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 $\scriptstyle {^{c}}\!$ (co-divisor 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 |
| (Co-divisor 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 | Resulting in | Resulting in |
| s.1 | Definition | $\scriptstyle {q{\tilde {\leftarrow }}p=q'p\,}\!$ | $\scriptstyle {q{\tilde {\leftarrow }}p=q'p\,}\!$ | $\scriptstyle {q{\tilde {\leftarrow }}p=q'p\,}\!$ |
| s.2 | Definition | $\scriptstyle {p{\tilde {\leftarrow }}q=p'q\,}\!$ | $\scriptstyle {p{\tilde {\leftarrow }}q=p'q\,}\!$ | $\scriptstyle {p{\tilde {\leftarrow }}q=p'q\,}\!$ |
| s.3 | s.1 s.2 | $\scriptstyle {q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q'p=qp'\,}\!$ | $\scriptstyle {q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q'p=qp'\,}\!$ | $\scriptstyle {q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q'p=qp'\,}\!$ |
| 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 | $\scriptstyle {q=qp+qp'\,}\!$ | $\scriptstyle {q=qp+qp'\,}\!$ | $\scriptstyle {q=qp+qp'\,}\!$ |
| 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 | $\scriptstyle {q'p=qp'\ \Rightarrow \ q=p\,}\!$ | $\scriptstyle {q'p=qp'\ \Rightarrow \ q=p\,}\!$ | $\scriptstyle {q'p=qp'\ \Rightarrow \ q=p\,}\!$ |
| s.13 | | $\scriptstyle {q=p\ \Rightarrow \ q'p=qp'\,}\!$ | $\scriptstyle {q=p\ \Rightarrow \ q'p=qp'\,}\!$ | $\scriptstyle {q=p\ \Rightarrow \ q'p=qp'\,}\!$ |
| s.14 | s.12 s.13 | $\scriptstyle {q=p\ \Leftrightarrow \ q'p=qp'\,}\!$ | $\scriptstyle {q=p\ \Leftrightarrow \ q'p=qp'\,}\!$ | $\scriptstyle {q=p\ \Leftrightarrow \ q'p=qp'\,}\!$ |
| s.15 | s.3 s.14 | $\scriptstyle {q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q=p\,}\!$ | $\scriptstyle {q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q=p\,}\!$ | $\scriptstyle {q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q=p\,}\!$ |
Implication is the dual of Converse Nonimplication :
| Step | Make use of | Resulting in | Resulting in | 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 | $\scriptstyle {\operatorname {dual} (q{\tilde {\leftarrow }}p)=q\rightarrow p\,}\!$ | $\scriptstyle {\operatorname {dual} (q{\tilde {\leftarrow }}p)=q\rightarrow p\,}\!$ | $\scriptstyle {\operatorname {dual} (q{\tilde {\leftarrow }}p)=q\rightarrow p\,}\!$ |

## 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.