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Is “p not implies q” logically equivalent to “not p implies q”?

compound proposition logically equivalentMaterial Conditional Not Logically Equivalent?Are these two statements logically equivalent?Logically equivalent formulas and contradictionProve that B is logically equivalent to C if and only if B logically implies C and C logically implies B.Truth table for logically equivalent questionsquantifiers in logically equivalent implicationsCan two propositions be both tautologies but not logically equivalent?Why are “$(P ⇒ Q) ⇒ R$” and “$P ⇒ (Q ⇒ R)$” not logically equivalent?

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4

$begingroup$

I am confused about whether $(A notrightarrow B)$ is logically equivalent to ~$(A rightarrow B)$?

logic propositional-calculus

share|cite|improve this question

asked 9 hours ago

Aashish Loknath PanigrahiAashish Loknath Panigrahi

25111 silver badge77 bronze badges

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• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
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  I have never seen "not implies" as a logical connective. Do you know its truth table?
  $endgroup$
  – Arturo Magidin
  9 hours ago
  

  
  
  
  


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  I haven't seen it either, where did you get this question from?, maybe a textbook?
  $endgroup$
  – Donlans Donlans
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  :( No I don't. I couldn't find any, that's why I got confused about whether "p not implies q" is equivalent to "not p implies q".
  $endgroup$
  – Aashish Loknath Panigrahi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Usually if you have any binary boolean operator $a?b$, then $anot?b$ means $lnot(a?b)$.
  $endgroup$
  – celtschk
  9 hours ago
  

  
  
  
  


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  @AashishLoknathPanigrahi: It looks like this could just be a case of awkward translation. Usually, we say "$p$ does not imply $q$", to mean precisely $neg(pto q)$. It could have been expressed (somewhat ungrammatical and non-standard) as "$p$ not implies $q$".
  $endgroup$
  – Arturo Magidin
  9 hours ago
  

  
  
  
  

 | 
show 6 more comments

4

$begingroup$

I am confused about whether $(A notrightarrow B)$ is logically equivalent to ~$(A rightarrow B)$?

logic propositional-calculus

share|cite|improve this question

asked 9 hours ago

Aashish Loknath PanigrahiAashish Loknath Panigrahi

25111 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  I have never seen "not implies" as a logical connective. Do you know its truth table?
  $endgroup$
  – Arturo Magidin
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I haven't seen it either, where did you get this question from?, maybe a textbook?
  $endgroup$
  – Donlans Donlans
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  :( No I don't. I couldn't find any, that's why I got confused about whether "p not implies q" is equivalent to "not p implies q".
  $endgroup$
  – Aashish Loknath Panigrahi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Usually if you have any binary boolean operator $a?b$, then $anot?b$ means $lnot(a?b)$.
  $endgroup$
  – celtschk
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @AashishLoknathPanigrahi: It looks like this could just be a case of awkward translation. Usually, we say "$p$ does not imply $q$", to mean precisely $neg(pto q)$. It could have been expressed (somewhat ungrammatical and non-standard) as "$p$ not implies $q$".
  $endgroup$
  – Arturo Magidin
  9 hours ago
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

I am confused about whether $(A notrightarrow B)$ is logically equivalent to ~$(A rightarrow B)$?

logic propositional-calculus

share|cite|improve this question

asked 9 hours ago

Aashish Loknath PanigrahiAashish Loknath Panigrahi

25111 silver badge77 bronze badges

$endgroup$

share|cite|improve this question

asked 9 hours ago

Aashish Loknath PanigrahiAashish Loknath Panigrahi

25111 silver badge77 bronze badges

share|cite|improve this question

asked 9 hours ago

Aashish Loknath PanigrahiAashish Loknath Panigrahi

25111 silver badge77 bronze badges

asked 9 hours ago

Aashish Loknath PanigrahiAashish Loknath Panigrahi

25111 silver badge77 bronze badges

asked 9 hours ago

Aashish Loknath PanigrahiAashish Loknath Panigrahi

25111 silver badge77 bronze badges

2 Answers
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5

$begingroup$

I would genreally consider $pnotto q$ to be an abbreviation for $neg(pto q)$, so they are equivalent by definition!

Note, however, that $neg(pto q)$ is not the same as $(neg p)to q$, and if you write $neg pto q$ it generally means the latter of there.

share|cite|improve this answer

answered 7 hours ago

Henning MakholmHenning Makholm

255k1818 gold badges336336 silver badges582582 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Based on a really brief Google search, I think you are correct. The vocabulary and notation is non-standard, but it appears to be useful if you wanted to create binary connectives for all 16 possible Boolean binary operators. I'll see if I can paste in a copy of some text that uses it.

ETA: check https://www.physicsforums.com/threads/unbounded-logical-trees.880905/ near the top. If someone can edit that tree into my answer as code, that would be sweet.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Matthew DalyMatthew Daly

3,55033 silver badges2525 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  A forum post of someone else asking a question is not a compelling source, especially someone who, to put it charitably, is not the clearest of communicators.
  $endgroup$
  – Derek Elkins
  6 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

I would genreally consider $pnotto q$ to be an abbreviation for $neg(pto q)$, so they are equivalent by definition!

Note, however, that $neg(pto q)$ is not the same as $(neg p)to q$, and if you write $neg pto q$ it generally means the latter of there.

share|cite|improve this answer

answered 7 hours ago

Henning MakholmHenning Makholm

255k1818 gold badges336336 silver badges582582 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

I would genreally consider $pnotto q$ to be an abbreviation for $neg(pto q)$, so they are equivalent by definition!

Note, however, that $neg(pto q)$ is not the same as $(neg p)to q$, and if you write $neg pto q$ it generally means the latter of there.

share|cite|improve this answer

answered 7 hours ago

Henning MakholmHenning Makholm

255k1818 gold badges336336 silver badges582582 bronze badges

$endgroup$

add a comment | 

$begingroup$

I would genreally consider $pnotto q$ to be an abbreviation for $neg(pto q)$, so they are equivalent by definition!

Note, however, that $neg(pto q)$ is not the same as $(neg p)to q$, and if you write $neg pto q$ it generally means the latter of there.

share|cite|improve this answer

answered 7 hours ago

Henning MakholmHenning Makholm

255k1818 gold badges336336 silver badges582582 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Henning MakholmHenning Makholm

255k1818 gold badges336336 silver badges582582 bronze badges

answered 7 hours ago

Henning MakholmHenning Makholm

255k1818 gold badges336336 silver badges582582 bronze badges

answered 7 hours ago

Henning MakholmHenning Makholm

255k1818 gold badges336336 silver badges582582 bronze badges

1

$begingroup$

Based on a really brief Google search, I think you are correct. The vocabulary and notation is non-standard, but it appears to be useful if you wanted to create binary connectives for all 16 possible Boolean binary operators. I'll see if I can paste in a copy of some text that uses it.

ETA: check https://www.physicsforums.com/threads/unbounded-logical-trees.880905/ near the top. If someone can edit that tree into my answer as code, that would be sweet.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Matthew DalyMatthew Daly

3,55033 silver badges2525 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  A forum post of someone else asking a question is not a compelling source, especially someone who, to put it charitably, is not the clearest of communicators.
  $endgroup$
  – Derek Elkins
  6 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Based on a really brief Google search, I think you are correct. The vocabulary and notation is non-standard, but it appears to be useful if you wanted to create binary connectives for all 16 possible Boolean binary operators. I'll see if I can paste in a copy of some text that uses it.

ETA: check https://www.physicsforums.com/threads/unbounded-logical-trees.880905/ near the top. If someone can edit that tree into my answer as code, that would be sweet.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Matthew DalyMatthew Daly

3,55033 silver badges2525 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  A forum post of someone else asking a question is not a compelling source, especially someone who, to put it charitably, is not the clearest of communicators.
  $endgroup$
  – Derek Elkins
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Based on a really brief Google search, I think you are correct. The vocabulary and notation is non-standard, but it appears to be useful if you wanted to create binary connectives for all 16 possible Boolean binary operators. I'll see if I can paste in a copy of some text that uses it.

ETA: check https://www.physicsforums.com/threads/unbounded-logical-trees.880905/ near the top. If someone can edit that tree into my answer as code, that would be sweet.

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Matthew DalyMatthew Daly

3,55033 silver badges2525 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Matthew DalyMatthew Daly

3,55033 silver badges2525 bronze badges

answered 9 hours ago

Matthew DalyMatthew Daly

3,55033 silver badges2525 bronze badges

answered 9 hours ago

Matthew DalyMatthew Daly

3,55033 silver badges2525 bronze badges