# Converse nonimplication

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

## Definition

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

### Truth table

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

p q
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).

Also related to the relative complement (set theory), where the relative complement of A in B is denoted B ∖ A.

## 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 ${\displaystyle \textstyle {p\not \subset q}}$ are

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

"not A but B"

## Boolean algebra

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

Example of a 2-element Boolean algebra: the 2 elements {0,1} with 0 as zero and 1 as unity element, operators ${\displaystyle \scriptstyle {\sim }\!}$ as complement operator, ${\displaystyle \scriptstyle {_{\vee }}\!}$ as join operator and ${\displaystyle \scriptstyle {_{\wedge }}\!}$ as meet operator, build the Boolean algebra of propositional logic.

 0 1 ${\displaystyle \scriptstyle {\sim x}\!}$ 1 0 x
and
1 0 y 1 1 0 1 ${\displaystyle \scriptstyle {y_{\vee }x}\!}$ x
and
1 0 y 0 1 0 0 ${\displaystyle \scriptstyle {y_{\wedge }x}\!}$ 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 ${\displaystyle \scriptstyle {dual(q{\tilde {\leftarrow }}p)\,}\!}$ ${\displaystyle \scriptstyle {=\,}\!}$ ${\displaystyle \scriptstyle {dual(q'p)\,}\!}$
${\displaystyle \scriptstyle \mathrm {s.2} }$ ${\displaystyle \scriptstyle \mathrm {s.1.right} }$ - .'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 ${\displaystyle \scriptstyle {=\,}\!}$ ${\displaystyle \scriptstyle {(p'q)'\,}\!}$
${\displaystyle \scriptstyle \mathrm {s.6} }$ ${\displaystyle \scriptstyle \mathrm {s.5.right} }$ ${\displaystyle \scriptstyle {=\,}\!}$ ${\displaystyle \scriptstyle {(p{\tilde {\leftarrow }}q)'\,}\!}$
${\displaystyle \scriptstyle \mathrm {s.7} }$ ${\displaystyle \scriptstyle \mathrm {s.6.right} }$ ${\displaystyle \scriptstyle {=\,}\!}$ ${\displaystyle \scriptstyle {p\leftarrow q\,}\!}$
${\displaystyle \scriptstyle \mathrm {s.8} }$ ${\displaystyle \scriptstyle \mathrm {s.7.right} }$ ${\displaystyle \scriptstyle {=\,}\!}$ ${\displaystyle \scriptstyle {q\rightarrow p\,}\!}$
${\displaystyle \scriptstyle \mathrm {s.9} }$ ${\displaystyle \scriptstyle \mathrm {s.1.left=s.8.right} }$ ${\displaystyle \scriptstyle {dual(q{\tilde {\leftarrow }}p)=q\rightarrow p\,}\!}$

## 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]