anonymous
  • anonymous
How can I simplify this: \[\left(P\wedge\neg R\right)\vee\left(P\wedge\neg Q\right)\vee\left(Q\wedge\neg R\right)\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I've been working on this but it seem I can't do anymore. =/
anonymous
  • anonymous
The original expression is this: \[\left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right]\]
anonymous
  • anonymous
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.

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anonymous
  • anonymous
This is the work I've done so far.
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abb0t
  • abb0t
Well, tautology is a proposition that is always true. That's all I know. That's all I know.
anonymous
  • anonymous
good reasoning @No-data
anonymous
  • anonymous
Thank you @Rohangrr
anonymous
  • anonymous
@brinethery Do you have any idea?
anonymous
  • anonymous
Are you sure it's true? It looks like the converse is true.
anonymous
  • anonymous
\[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]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
  • anonymous
I feel like q is irrelvent and its just If P then R
anonymous
  • anonymous
\[ \left(P\Rightarrow R\right)\Rightarrow\left[\left(P\Rightarrow Q\right)\wedge\left(Q\Rightarrow R\right)\right] \]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
  • anonymous
^^ makes alot of sense
anonymous
  • anonymous
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.
anonymous
  • anonymous
I did not say it was true
anonymous
  • anonymous
Mmm I don't understand the expression after "Suppose" what those arrows mean?
anonymous
  • anonymous
Ok I got the counter example, is a row on my truth table and its correct. Thank you @wio
anonymous
  • anonymous
I was just trying to show that the expression is not a tautology.
anonymous
  • anonymous
The expression you wrote saying you thought it should be is, in fact, a tautology.
anonymous
  • anonymous
Imagine if I have you the proposition \(P\) and claimed it was a tautology. How would you prove it is wrong?
anonymous
  • anonymous
There is no boolean algebra to prove it, you just give a counter example, right?
anonymous
  • anonymous
Do you think that is not possible to find a way to show this using only theorems and definitions?
anonymous
  • anonymous
Which theorems?
anonymous
  • anonymous
How would you show that \(P\) in and of itself is not a tautology?
anonymous
  • anonymous
There is no way to know. Unless you tell me that P can take the values T and F.
anonymous
  • anonymous
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.
anonymous
  • anonymous
and then, by definition of P. P is not a tautology
anonymous
  • anonymous
So you'd say by definition of proposition, a proposition is not a tautology? Then how about \(P\wedge Q\)?
anonymous
  • anonymous
You meant that proposition is not a tautology.
anonymous
  • anonymous
As far as I know a tautology is a predicate that is always true for any value of their parameters.
anonymous
  • anonymous
\[P\wedge Q\] is a predicate that can be true or false. So it's not a tautology.
anonymous
  • anonymous
Anyway I appreciate your help.
anonymous
  • anonymous
You're claiming it can be false... The point is to get you to prove it without counterexample.
anonymous
  • anonymous
I'm just curious as to the rules of the game.
anonymous
  • anonymous
I think I'm not using a counter example and that is a valid proof.
anonymous
  • anonymous
A tautology is a restatement within the same premiss like all trees are made of wood
anonymous
  • anonymous
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?
anonymous
  • anonymous
I don't think saying "is a predicate that can be true or false." is valid without a counter example.
anonymous
  • anonymous
Or maybe there is another way, but I'm not sure
anonymous
  • anonymous
Is not evident that that predicate can be true or false. Its not a valid proof.
anonymous
  • anonymous
and \[P\wedge Q\]is by definition true or false.
anonymous
  • anonymous
Sure but then what makes something 'evident'?
anonymous
  • anonymous
Is there some standard form you want it to be in?
anonymous
  • anonymous
I wanted to say that there is a definition by saying "evident"
anonymous
  • anonymous
What about \(P\wedge Q \wedge R\)?
anonymous
  • anonymous
I accepted you method buy I feel like you're trying to say that there is no other method to prove this problem.
anonymous
  • anonymous
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.
anonymous
  • anonymous
let \[P\wedge Q\Leftrightarrow S \]\[S\wedge R\]
anonymous
  • anonymous
I mean prove without counter example
anonymous
  • anonymous
so it's not a tautology.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Using your logic that \(\wedge\) doesn't give a tautology, then couldn't we say \(\neg P\vee P\) isn't a tautology?
anonymous
  • anonymous
ITs a Syllogism if the expanded form is the antecedent, but because its the consequent it is not tautological
anonymous
  • anonymous
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.
anonymous
  • anonymous
q's relation to p and r
anonymous
  • anonymous
This is not a tautology:\[P\wedge Q\], not just the \[\wedge \] symbol.
anonymous
  • anonymous
Sorry I did not understand that @Edutopia
anonymous
  • anonymous
I'm just beginning with this logic things lol.
anonymous
  • anonymous
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
anonymous
  • anonymous
ahh that is right Edutopia.
anonymous
  • anonymous
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".
anonymous
  • anonymous
But maybe that is a semantic argument? It's just my thought.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Yeah it's interesting @wio and I agree your counterexamples are a good tool. I just think they are not elegant.
anonymous
  • anonymous
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 \(Q\vee \neg Q\).
anonymous
  • anonymous
I'm not sure, but I think it's valid in math.If\[x=y\]then\[x+a=y+a\]
anonymous
  • anonymous
That is the addition property of equality. \(Q\) was not being appended to both sides of the implication.
anonymous
  • anonymous
yes, but in math its all a valid argument because you are using numbers
anonymous
  • anonymous
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.
anonymous
  • anonymous
a=a and so forth
anonymous
  • anonymous
@No-data I have no idea :-). I'm so stupid.
anonymous
  • anonymous
Is this for linear algebra?
anonymous
  • anonymous
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.

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