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

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 ↚ Q {\displaystyle P\nleftarrow Q}, or P ⊄ Q {\displaystyle P\not \subset Q}, and is logically equivalent to ¬ ( P ← Q ) {\displaystyle \neg (P\leftarrow Q)} and ¬ P ∧ Q {\displaystyle \neg P\wedge Q}.

Truth table

The truth table of A ↚ B {\displaystyle A\nleftarrow B}.

A {\displaystyle A}B {\displaystyle B}A ↚ B {\displaystyle A\nleftarrow B}
FFF
FTT
TFF
TTF

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

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

Not P, but Q.

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 1andy 1 1 1 0 0 1 y ∨ x {\textstyle y_{\vee }x} 0 1 xandy 1 0 1 0 0 0 y ∧ x {\textstyle y_{\wedge }x} 0 1 xthen y ↚ x {\displaystyle \scriptstyle {y\nleftarrow x}\!} meansy 1 0 0 0 0 1 y ↚ x {\displaystyle \scriptstyle {y\nleftarrow x}\!} 0 1 x
∼ x {\textstyle {}\sim x}10
x01
y
111
001
y ∨ x {\textstyle y_{\vee }x}01x
y
101
000
y ∧ x {\textstyle y_{\wedge }x}01x
y
100
001
y ↚ x {\displaystyle \scriptstyle {y\nleftarrow x}\!}01x
(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 {\displaystyle \scriptstyle {^{c}}\!} (co-divisor 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.

x c {\displaystyle \scriptstyle {x^{c}}\!} 6 3 2 1 x 1 2 3 6andy 6 6 6 6 6 3 3 6 3 6 2 2 2 6 6 1 1 2 3 6 y ∨ x {\displaystyle \scriptstyle {y_{\vee }x}\!} 1 2 3 6 xandy 6 1 2 3 6 3 1 1 3 3 2 1 2 1 2 1 1 1 1 1 y ∧ x {\displaystyle \scriptstyle {y_{\wedge }x}} 1 2 3 6 xthen y ↚ x {\displaystyle \scriptstyle {y\nleftarrow x}\!} meansy 6 1 1 1 1 3 1 2 1 2 2 1 1 3 3 1 1 2 3 6 y ↚ x {\displaystyle \scriptstyle {y\nleftarrow x}\!} 1 2 3 6 x
x c {\displaystyle \scriptstyle {x^{c}}\!}6321
x1236
y
66666
33636
22266
11236
y ∨ x {\displaystyle \scriptstyle {y_{\vee }x}\!}1236x
y
61236
31133
21212
11111
y ∧ x {\displaystyle \scriptstyle {y_{\wedge }x}}1236x
y
61111
31212
21133
11236
y ↚ x {\displaystyle \scriptstyle {y\nleftarrow x}\!}1236x
(Co-divisor 6)(Least common multiple)(Greatest common divisor)(x's greatest divisor coprime with y)

Properties

Non-associative

r ↚ ( q ↚ p ) = ( r ↚ q ) ↚ p {\displaystyle r\nleftarrow (q\nleftarrow p)=(r\nleftarrow q)\nleftarrow p} if and only if r p = 0 {\displaystyle rp=0} #s5 (In a two-element Boolean algebra the latter condition is reduced to r = 0 {\displaystyle r=0} or p = 0 {\displaystyle p=0}). Hence in a nontrivial Boolean algebra converse nonimplication is nonassociative. ( r ↚ q ) ↚ p = r ′ q ↚ p (by definition) = ( r ′ q ) ′ p (by definition) = ( r + q ′ ) p (De Morgan's laws) = ( r + r ′ q ′ ) p (Absorption law) = r p + r ′ q ′ p = r p + r ′ ( q ↚ p ) (by definition) = r p + r ↚ ( q ↚ p ) (by definition) {\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 if and only if r p = 0 {\displaystyle rp=0}.

Non-commutative

  • q ↚ p = p ↚ q {\displaystyle q\nleftarrow p=p\nleftarrow q} if and only if q = p {\displaystyle q=p} #s6. Hence converse nonimplication is noncommutative.

Neutral and absorbing elements

  • 0 is a left neutral element (0 ↚ p = p {\displaystyle 0\nleftarrow p=p}) and a right absorbing element (p ↚ 0 = 0 {\displaystyle {p\nleftarrow 0=0}}).
  • 1 ↚ p = 0 {\displaystyle 1\nleftarrow p=0}, p ↚ 1 = p ′ {\displaystyle p\nleftarrow 1=p'}, and p ↚ p = 0 {\displaystyle p\nleftarrow p=0}.
  • Implication q → p {\displaystyle q\rightarrow p} is the dual of converse nonimplication q ↚ p {\displaystyle q\nleftarrow p} #s7.
Converse nonimplication is noncommutative
StepMake use ofResulting in
s.1Definitionq ← ~ p = q ′ p {\displaystyle \scriptstyle {q{\tilde {\leftarrow }}p=q'p\,}\!}
s.2Definitionp ← ~ q = p ′ q {\displaystyle \scriptstyle {p{\tilde {\leftarrow }}q=p'q\,}\!}
s.3s.1 s.2q ← ~ p = p ← ~ q ⇔ q ′ p = q p ′ {\displaystyle \scriptstyle {q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q'p=qp'\,}\!}
s.4q {\displaystyle \scriptstyle {q\,}\!}= {\displaystyle \scriptstyle {=\,}\!}q .1 {\displaystyle \scriptstyle {q.1\,}\!}
s.5s.4.right - expand Unit element= {\displaystyle \scriptstyle {=\,}\!}q . ( p + p ′ ) {\displaystyle \scriptstyle {q.(p+p')\,}\!}
s.6s.5.right - evaluate expression= {\displaystyle \scriptstyle {=\,}\!}q p + q p ′ {\displaystyle \scriptstyle {qp+qp'\,}\!}
s.7s.4.left = s.6.rightq = q p + q p ′ {\displaystyle \scriptstyle {q=qp+qp'\,}\!}
s.8q ′ p = q p ′ {\displaystyle \scriptstyle {q'p=qp'\,}\!}⇒ {\displaystyle \scriptstyle {\Rightarrow \,}\!}q p + q p ′ = q p + q ′ p {\displaystyle \scriptstyle {qp+qp'=qp+q'p\,}\!}
s.9s.8 - regroup common factors⇒ {\displaystyle \scriptstyle {\Rightarrow \,}\!}q . ( p + p ′ ) = ( q + q ′ ) . p {\displaystyle \scriptstyle {q.(p+p')=(q+q').p\,}\!}
s.10s.9 - join of complements equals unity⇒ {\displaystyle \scriptstyle {\Rightarrow \,}\!}q .1 = 1. p {\displaystyle \scriptstyle {q.1=1.p\,}\!}
s.11s.10.right - evaluate expression⇒ {\displaystyle \scriptstyle {\Rightarrow \,}\!}q = p {\displaystyle \scriptstyle {q=p\,}\!}
s.12s.8 s.11q ′ p = q p ′ ⇒ q = p {\displaystyle \scriptstyle {q'p=qp'\ \Rightarrow \ q=p\,}\!}
s.13q = p ⇒ q ′ p = q p ′ {\displaystyle \scriptstyle {q=p\ \Rightarrow \ q'p=qp'\,}\!}
s.14s.12 s.13q = p ⇔ q ′ p = q p ′ {\displaystyle \scriptstyle {q=p\ \Leftrightarrow \ q'p=qp'\,}\!}
s.15s.3 s.14q ← ~ p = p ← ~ q ⇔ q = p {\displaystyle \scriptstyle {q{\tilde {\leftarrow }}p=p{\tilde {\leftarrow }}q\ \Leftrightarrow \ q=p\,}\!}
Implication is the dual of converse nonimplication
StepMake use ofResulting in
s.1Definitiondual ⁡ ( q ← ~ p ) {\displaystyle \scriptstyle {\operatorname {dual} (q{\tilde {\leftarrow }}p)\,}\!}= {\displaystyle \scriptstyle {=\,}\!}dual ⁡ ( q ′ p ) {\displaystyle \scriptstyle {\operatorname {dual} (q'p)\,}\!}
s.2s.1.right - .'s dual is += {\displaystyle \scriptstyle {=\,}\!}q ′ + p {\displaystyle \scriptstyle {q'+p\,}\!}
s.3s.2.right - Involution complement= {\displaystyle \scriptstyle {=\,}\!}( q ′ + p ) ″ {\displaystyle \scriptstyle {(q'+p)''\,}\!}
s.4s.3.right - De Morgan's laws applied once= {\displaystyle \scriptstyle {=\,}\!}( q p ′ ) ′ {\displaystyle \scriptstyle {(qp')'\,}\!}
s.5s.4.right - Commutative law= {\displaystyle \scriptstyle {=\,}\!}( p ′ q ) ′ {\displaystyle \scriptstyle {(p'q)'\,}\!}
s.6s.5.right= {\displaystyle \scriptstyle {=\,}\!}( p ← ~ q ) ′ {\displaystyle \scriptstyle {(p{\tilde {\leftarrow }}q)'\,}\!}
s.7s.6.right= {\displaystyle \scriptstyle {=\,}\!}p ← q {\displaystyle \scriptstyle {p\leftarrow q\,}\!}
s.8s.7.right= {\displaystyle \scriptstyle {=\,}\!}q → p {\displaystyle \scriptstyle {q\rightarrow p\,}\!}
s.9s.1.left = s.8.rightdual ⁡ ( q ← ~ p ) = q → p {\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.

External links

  • Media related toConverse nonimplicationat Wikimedia Commons