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anonymous
 3 years ago
How can I simplify this: \[\left(P\wedge\neg R\right)\vee\left(P\wedge\neg Q\right)\vee\left(Q\wedge\neg R\right)\]
anonymous
 3 years ago
How can I simplify this: \[\left(P\wedge\neg R\right)\vee\left(P\wedge\neg Q\right)\vee\left(Q\wedge\neg R\right)\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I've been working on this but it seem I can't do anymore. =/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The original expression is this: \[\left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I see that it is like prove the transitivy propertie for the implication operator. I made a truth table for this and it turned out to be false, but I want to get to the same conclussion without using truth tables.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is the work I've done so far.

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0Well, tautology is a proposition that is always true. That's all I know. That's all I know.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0good reasoning @Nodata

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@brinethery Do you have any idea?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are you sure it's true? It looks like the converse is true.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]Suppose \(P \gets T\quad Q\gets F\quad R\gets T \) \[ (T\implies T)\implies [(T\implies F)\wedge (F \implies T)] \\ T\implies (F\wedge T) \\ T\implies F \\ F \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I feel like q is irrelvent and its just If P then R

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]I think it should be \[ \left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \Rightarrow \left(P\Rightarrow R\right) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0^^ makes alot of sense

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There is nothing wrong with using a counter example to show something is false. Trying to prove it without a counter example seems pointless to me.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did not say it was true

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Mmm I don't understand the expression after "Suppose" what those arrows mean?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok I got the counter example, is a row on my truth table and its correct. Thank you @wio

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I was just trying to show that the expression is not a tautology.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The expression you wrote saying you thought it should be is, in fact, a tautology.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Imagine if I have you the proposition \(P\) and claimed it was a tautology. How would you prove it is wrong?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There is no boolean algebra to prove it, you just give a counter example, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you think that is not possible to find a way to show this using only theorems and definitions?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How would you show that \(P\) in and of itself is not a tautology?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There is no way to know. Unless you tell me that P can take the values T and F.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I mean, I have nothing against trying to do things a particular way, but what are the rules really? How would you show \(P\) is not a tautology without a counter example.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and then, by definition of P. P is not a tautology

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you'd say by definition of proposition, a proposition is not a tautology? Then how about \(P\wedge Q\)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You meant that proposition is not a tautology.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0As far as I know a tautology is a predicate that is always true for any value of their parameters.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[P\wedge Q\] is a predicate that can be true or false. So it's not a tautology.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Anyway I appreciate your help.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You're claiming it can be false... The point is to get you to prove it without counterexample.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm just curious as to the rules of the game.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think I'm not using a counter example and that is a valid proof.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A tautology is a restatement within the same premiss like all trees are made of wood

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What if I say "is a predicate that can be true or false.", about the original predicate? Said it about \[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]Would simply saying that be a valid proof that it is not a tautology?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't think saying "is a predicate that can be true or false." is valid without a counter example.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Or maybe there is another way, but I'm not sure

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is not evident that that predicate can be true or false. Its not a valid proof.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and \[P\wedge Q\]is by definition true or false.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sure but then what makes something 'evident'?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is there some standard form you want it to be in?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I wanted to say that there is a definition by saying "evident"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What about \(P\wedge Q \wedge R\)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I accepted you method buy I feel like you're trying to say that there is no other method to prove this problem.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What I am saying is that if there is another way to prove this problem, then there needs to be a way to prove very simple predicates.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let \[P\wedge Q\Leftrightarrow S \]\[S\wedge R\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I mean prove without counter example

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it's not a tautology.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know that, I already proved it using other method. I just wanted to know if it was possible to do it the way I'm trying.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Using your logic that \(\wedge\) doesn't give a tautology, then couldn't we say \(\neg P\vee P\) isn't a tautology?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ITs a Syllogism if the expanded form is the antecedent, but because its the consequent it is not tautological

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well it's just an interesting thing to think about. I would like to find a way to prove false without counterexample but it seems hard to understand what is allowed.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0q's relation to p and r

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is not a tautology:\[P\wedge Q\], not just the \[\wedge \] symbol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry I did not understand that @Edutopia

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm just beginning with this logic things lol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my comp is sllooow, i ment to say: q's relation to p and r would have to be in the premiss for it to be tautological

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ahh that is right Edutopia.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You know what's interesting about proving something is true, is that you are showing it is true for all cases. If you are proving something is false, you only need to show one case is false. To prove that it is false for all cases can't always be done and doesn't need to be done. I think to even say \(P\) is not a tautology, you are implicitly calling upon the counter example just to claim "\(P\) can be false".

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But maybe that is a semantic argument? It's just my thought.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in IF (IF P THEN R) THEN BOTH (IF P THEN Q) AND ( IF Q THEN R) the relation of Q to P and R is not established, imagine the argument where P and R have to do with Physics and Q is someones opinion on abortion.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah it's interesting @wio and I agree your counterexamples are a good tool. I just think they are not elegant.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When I first saw \[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]The \(Q\) made me think it's likely not a tautology right away. Since it wasn't something inconsequential like \(Q\vee \neg Q\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not sure, but I think it's valid in math.If\[x=y\]then\[x+a=y+a\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is the addition property of equality. \(Q\) was not being appended to both sides of the implication.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, but in math its all a valid argument because you are using numbers

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know it just look similar. Thank you guys. It is nice to be able to discuss this kind of things with other people. Thank you for you r time.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Nodata I have no idea :). I'm so stupid.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Is this for linear algebra?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It is propositional logic. You're not stupid at all @brinethery. I was curious about how databases work so I picked a book about math applied to databases and this is a problem from the first chapter. Logic and Set Theory are the foundation of DB systems.
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