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 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.
May 17th, 2018, 08:09 PM   #2
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 Originally Posted by moredumbimpossible 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.

May 17th, 2018, 08:34 PM   #3
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 Originally Posted by Maschke 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.?

 May 17th, 2018, 08:58 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1122 Math Focus: Elementary mathematics and beyond I'm looking at [(p ∧ q) ↓ q] $\implies$ ¬q, aren't I? That's true, isn't it?
May 17th, 2018, 09:19 PM   #5
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 Originally Posted by greg1313 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

 May 17th, 2018, 09:27 PM #6 Senior Member     Joined: Sep 2015 From: USA Posts: 2,369 Thanks: 1273 \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
 May 17th, 2018, 09:32 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1122 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.
May 17th, 2018, 09:34 PM   #8
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 Originally Posted by romsek \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.

May 17th, 2018, 10:04 PM   #9
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 Originally Posted by greg1313 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.

May 18th, 2018, 12:50 AM   #10
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 Originally Posted by greg1313 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.

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