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

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NodataBest ResponseYou've already chosen the best response.2
I've been working on this but it seem I can't do anymore. =/
 one year ago

NodataBest ResponseYou've already chosen the best response.2
The original expression is this: \[\left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right]\]
 one year ago

NodataBest ResponseYou've already chosen the best response.2
I 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.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
This is the work I've done so far.
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
Well, tautology is a proposition that is always true. That's all I know. That's all I know.
 one year ago

RohangrrBest ResponseYou've already chosen the best response.1
good reasoning @Nodata
 one year ago

NodataBest ResponseYou've already chosen the best response.2
@brinethery Do you have any idea?
 one year ago

wioBest ResponseYou've already chosen the best response.2
Are you sure it's true? It looks like the converse is true.
 one year ago

wioBest ResponseYou've already chosen the best response.2
\[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]Suppose \(P \gets T\quad Q\gets F\quad R\gets T \) \[ (T\implies T)\implies [(T\implies F)\wedge (F \implies T)] \\ T\implies (F\wedge T) \\ T\implies F \\ F \]
 one year ago

EdutopiaBest ResponseYou've already chosen the best response.0
I feel like q is irrelvent and its just If P then R
 one year ago

wioBest ResponseYou've already chosen the best response.2
\[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]I think it should be \[ \left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \Rightarrow \left(P\Rightarrow R\right) \]
 one year ago

EdutopiaBest ResponseYou've already chosen the best response.0
^^ makes alot of sense
 one year ago

wioBest ResponseYou've already chosen the best response.2
There 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.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
I did not say it was true
 one year ago

NodataBest ResponseYou've already chosen the best response.2
Mmm I don't understand the expression after "Suppose" what those arrows mean?
 one year ago

NodataBest ResponseYou've already chosen the best response.2
Ok I got the counter example, is a row on my truth table and its correct. Thank you @wio
 one year ago

NodataBest ResponseYou've already chosen the best response.2
I was just trying to show that the expression is not a tautology.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
The expression you wrote saying you thought it should be is, in fact, a tautology.
 one year ago

wioBest ResponseYou've already chosen the best response.2
Imagine if I have you the proposition \(P\) and claimed it was a tautology. How would you prove it is wrong?
 one year ago

wioBest ResponseYou've already chosen the best response.2
There is no boolean algebra to prove it, you just give a counter example, right?
 one year ago

NodataBest ResponseYou've already chosen the best response.2
Do you think that is not possible to find a way to show this using only theorems and definitions?
 one year ago

wioBest ResponseYou've already chosen the best response.2
How would you show that \(P\) in and of itself is not a tautology?
 one year ago

NodataBest ResponseYou've already chosen the best response.2
There is no way to know. Unless you tell me that P can take the values T and F.
 one year ago

wioBest ResponseYou've already chosen the best response.2
I 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.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
and then, by definition of P. P is not a tautology
 one year ago

wioBest ResponseYou've already chosen the best response.2
So you'd say by definition of proposition, a proposition is not a tautology? Then how about \(P\wedge Q\)?
 one year ago

NodataBest ResponseYou've already chosen the best response.2
You meant that proposition is not a tautology.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
As far as I know a tautology is a predicate that is always true for any value of their parameters.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
\[P\wedge Q\] is a predicate that can be true or false. So it's not a tautology.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
Anyway I appreciate your help.
 one year ago

wioBest ResponseYou've already chosen the best response.2
You're claiming it can be false... The point is to get you to prove it without counterexample.
 one year ago

wioBest ResponseYou've already chosen the best response.2
I'm just curious as to the rules of the game.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
I think I'm not using a counter example and that is a valid proof.
 one year ago

EdutopiaBest ResponseYou've already chosen the best response.0
A tautology is a restatement within the same premiss like all trees are made of wood
 one year ago

wioBest ResponseYou've already chosen the best response.2
What 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?
 one year ago

wioBest ResponseYou've already chosen the best response.2
I don't think saying "is a predicate that can be true or false." is valid without a counter example.
 one year ago

wioBest ResponseYou've already chosen the best response.2
Or maybe there is another way, but I'm not sure
 one year ago

NodataBest ResponseYou've already chosen the best response.2
Is not evident that that predicate can be true or false. Its not a valid proof.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
and \[P\wedge Q\]is by definition true or false.
 one year ago

wioBest ResponseYou've already chosen the best response.2
Sure but then what makes something 'evident'?
 one year ago

wioBest ResponseYou've already chosen the best response.2
Is there some standard form you want it to be in?
 one year ago

NodataBest ResponseYou've already chosen the best response.2
I wanted to say that there is a definition by saying "evident"
 one year ago

wioBest ResponseYou've already chosen the best response.2
What about \(P\wedge Q \wedge R\)?
 one year ago

NodataBest ResponseYou've already chosen the best response.2
I accepted you method buy I feel like you're trying to say that there is no other method to prove this problem.
 one year ago

wioBest ResponseYou've already chosen the best response.2
What 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.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
let \[P\wedge Q\Leftrightarrow S \]\[S\wedge R\]
 one year ago

wioBest ResponseYou've already chosen the best response.2
I mean prove without counter example
 one year ago

NodataBest ResponseYou've already chosen the best response.2
so it's not a tautology.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
I 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.
 one year ago

wioBest ResponseYou've already chosen the best response.2
Using your logic that \(\wedge\) doesn't give a tautology, then couldn't we say \(\neg P\vee P\) isn't a tautology?
 one year ago

EdutopiaBest ResponseYou've already chosen the best response.0
ITs a Syllogism if the expanded form is the antecedent, but because its the consequent it is not tautological
 one year ago

wioBest ResponseYou've already chosen the best response.2
Well 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.
 one year ago

EdutopiaBest ResponseYou've already chosen the best response.0
q's relation to p and r
 one year ago

NodataBest ResponseYou've already chosen the best response.2
This is not a tautology:\[P\wedge Q\], not just the \[\wedge \] symbol.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
Sorry I did not understand that @Edutopia
 one year ago

NodataBest ResponseYou've already chosen the best response.2
I'm just beginning with this logic things lol.
 one year ago

EdutopiaBest ResponseYou've already chosen the best response.0
my 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
 one year ago

NodataBest ResponseYou've already chosen the best response.2
ahh that is right Edutopia.
 one year ago

wioBest ResponseYou've already chosen the best response.2
You 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".
 one year ago

wioBest ResponseYou've already chosen the best response.2
But maybe that is a semantic argument? It's just my thought.
 one year ago

EdutopiaBest ResponseYou've already chosen the best response.0
in 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.
 one year ago

NodataBest ResponseYou've already chosen the best response.2
Yeah it's interesting @wio and I agree your counterexamples are a good tool. I just think they are not elegant.
 one year ago

wioBest ResponseYou've already chosen the best response.2
When 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 \(Q\vee \neg Q\).
 one year ago

NodataBest ResponseYou've already chosen the best response.2
I'm not sure, but I think it's valid in math.If\[x=y\]then\[x+a=y+a\]
 one year ago

wioBest ResponseYou've already chosen the best response.2
That is the addition property of equality. \(Q\) was not being appended to both sides of the implication.
 one year ago

EdutopiaBest ResponseYou've already chosen the best response.0
yes, but in math its all a valid argument because you are using numbers
 one year ago

NodataBest ResponseYou've already chosen the best response.2
I 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.
 one year ago

brinetheryBest ResponseYou've already chosen the best response.0
@Nodata I have no idea :). I'm so stupid.
 one year ago

brinetheryBest ResponseYou've already chosen the best response.0
Is this for linear algebra?
 one year ago

NodataBest ResponseYou've already chosen the best response.2
It 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.
 one year ago
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