My Math Forum  

Go Back   My Math Forum > Science Forums > Computer Science

Computer Science Computer Science Forum


Thanks Tree1Thanks
  • 1 Post By romsek
Reply
 
LinkBack Thread Tools Display Modes
May 17th, 2018, 08:01 PM   #1
Newbie
 
Joined: May 2018
From: notimportant

Posts: 9
Thanks: 0

Propositional Logic

Please help me with this thing. I'm so frustrated I can't understand propositional logic

Demonstrate this:

(p ∧ q) ↓ q ≡ ¬q

PLEASE.
moredumbimpossible is offline  
 
May 17th, 2018, 08:09 PM   #2
Senior Member
 
Joined: Aug 2012

Posts: 1,960
Thanks: 547

Quote:
Originally Posted by moredumbimpossible View Post
Please help me with this thing. I'm so frustrated I can't understand propositional logic

Demonstrate this:

(p ∧ q) ↓ q ≡ ¬q

PLEASE.
I had to look up the down arrow. That's logical nor. That is, $P \downarrow Q$ is true just in case both $P$ and $Q$ are false.

However with that interpretation of down arrow, your statement is not true.

On the right you have $q \equiv \neg q$. That is tautologically false.

On the left, $p \land q$ is true just in case both $p$ and $q$ are true.

So if $p$ is true and $q$ is true, then the left hand side of the down arrow is true, and the right hand side is false. So the down arrow expression is false.

Unless you are interpreting the down arrow with a different meaning, this is how I see it.
Maschke is offline  
May 17th, 2018, 08:34 PM   #3
Newbie
 
Joined: May 2018
From: notimportant

Posts: 9
Thanks: 0

Quote:
Originally Posted by Maschke View Post
I had to look up the down arrow. That's logical nor. That is, $P \downarrow Q$ is true just in case both $P$ and $Q$ are false.

However with that interpretation of down arrow, your statement is not true.

On the right you have $q \equiv \neg q$. That is tautologically false.

On the left, $p \land q$ is true just in case both $p$ and $q$ are true.

So if $p$ is true and $q$ is true, then the left hand side of the down arrow is true, and the right hand side is false. So the down arrow expression is false.

Unless you are interpreting the down arrow with a different meaning, this is how I see it.

Thanks., maybe you can solve that with the laws of logic.?
moredumbimpossible is offline  
May 17th, 2018, 08:58 PM   #4
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,825
Thanks: 1051

Math Focus: Elementary mathematics and beyond
I'm looking at [(p ∧ q) ↓ q] $\implies$ ¬q, aren't I? That's true, isn't it?
greg1313 is offline  
May 17th, 2018, 09:19 PM   #5
Newbie
 
Joined: May 2018
From: notimportant

Posts: 9
Thanks: 0

Quote:
Originally Posted by greg1313 View Post
I'm looking at [(p ∧ q) ↓ q] $\implies$ ¬q, aren't I? That's true, isn't it?
I'm not sure. But this problem needs to be solved/reduced applying the laws of logic.

those three lines mean "equal" or =
not the right arrow
moredumbimpossible is offline  
May 17th, 2018, 09:27 PM   #6
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 2,011
Thanks: 1044

$\begin{align*}

&(p \wedge q) \downarrow q \equiv \\ \\

&\neg((p\wedge q) \vee q) \equiv \\ \\

&\neg q

\end{align*}$

The last from the fact that $(p \wedge q) \subseteq q$
Thanks from greg1313
romsek is offline  
May 17th, 2018, 09:32 PM   #7
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,825
Thanks: 1051

Math Focus: Elementary mathematics and beyond
The three lines mean "logically equivalent to". That is, the statement to the left of the three lines is logically equivalent to the statement to the right of the three lines. If your problem was interpreted that way, it would appear valid.
greg1313 is offline  
May 17th, 2018, 09:34 PM   #8
Newbie
 
Joined: May 2018
From: notimportant

Posts: 9
Thanks: 0

Quote:
Originally Posted by romsek View Post
$\begin{align*}

&(p \wedge q) \downarrow q \equiv \\ \\

&\neg((p\wedge q) \vee q) \equiv \\ \\

&\neg q

\end{align*}$

The last from the fact that $(p \wedge q) \subseteq q$
Thanks dude.
moredumbimpossible is offline  
May 17th, 2018, 10:04 PM   #9
Newbie
 
Joined: May 2018
From: notimportant

Posts: 9
Thanks: 0

Quote:
Originally Posted by greg1313 View Post
The three lines mean "logically equivalent to". That is, the statement to the left of the three lines is logically equivalent to the statement to the right of the three lines. If your problem was interpreted that way, it would appear valid.
All I have is (p ∧ q) ↓ q ≡ ¬q and tells me to reduce using laws of logic, Morgan, double negation, distribution, etc..
Thanks.
moredumbimpossible is offline  
May 18th, 2018, 12:50 AM   #10
Senior Member
 
Joined: Aug 2012

Posts: 1,960
Thanks: 547

Quote:
Originally Posted by greg1313 View Post
I'm looking at [(p ∧ q) ↓ q] $\implies$ ¬q, aren't I? That's true, isn't it?
Oh I see. Good catch.

Wiki shows the order of precedence of the logical operators. Downarrow is not listed.

https://en.wikipedia.org/wiki/Logica..._of_precedence

Logical equivalence has the lowest precedence of all operators, so if we extend this principle to nor then the proposed solution is correct, that is we evaluate downarrow before equivalence.

I Googled around and could not find any definitive resolution to this question.

I found one Stackoverflow thread that notes that if an operator's precedence is not defined, it should be processed left to right and the proposed solution is correct. Still, this is all pretty murky. It depends on the order of precedence of an operator so obscure that its order of precedence is not explicitly defined anywhere on the Internet. One has to apply the principle that logical equivalence is always lower than any other operator anyone can think of.

https://stackoverflow.com/questions/...-nand-nor-xnor

Last edited by Maschke; May 18th, 2018 at 01:12 AM.
Maschke is offline  
Reply

  My Math Forum > Science Forums > Computer Science

Tags
logic, propositional



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Help Logic Propositional Calculus Bodner Abstract Algebra 0 September 16th, 2014 03:01 AM
Propositional logic question RyanPowers Computer Science 3 August 28th, 2014 06:16 AM
Propositional Logic - Is my answer correct? mohitpd Applied Math 0 October 26th, 2013 10:04 PM
Propositional Logic summer90 Applied Math 2 January 27th, 2010 09:26 AM
Propositional/Symbolic Logic help please ElMarsh Applied Math 27 April 9th, 2009 05:46 AM





Copyright © 2018 My Math Forum. All rights reserved.