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

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- schrodinger

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- anonymous

I've been working on this but it seem I can't do anymore. =/

- 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

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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## More answers

- anonymous

This is the work I've done so far.

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- abb0t

Well, tautology is a proposition that is always true. That's all I know. That's all I know.

- anonymous

good reasoning @No-data

- anonymous

Thank you @Rohangrr

- anonymous

@brinethery Do you have any idea?

- anonymous

Are you sure it's true? It looks like the converse is true.

- 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

I feel like q is irrelvent and its just If P then R

- 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

^^ makes alot of sense

- 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

I did not say it was true

- anonymous

Mmm I don't understand the expression after "Suppose" what those arrows mean?

- anonymous

Ok I got the counter example, is a row on my truth table and its correct. Thank you @wio

- anonymous

I was just trying to show that the expression is not a tautology.

- anonymous

The expression you wrote saying you thought it should be is, in fact, a tautology.

- anonymous

Imagine if I have you the proposition \(P\) and claimed it was a tautology. How would you prove it is wrong?

- anonymous

There is no boolean algebra to prove it, you just give a counter example, right?

- anonymous

Do you think that is not possible to find a way to show this using only theorems and definitions?

- anonymous

Which theorems?

- anonymous

How would you show that \(P\) in and of itself is not a tautology?

- anonymous

There is no way to know. Unless you tell me that P can take the values T and F.

- 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

and then, by definition of P. P is not a tautology

- anonymous

So you'd say by definition of proposition, a proposition is not a tautology? Then how about \(P\wedge Q\)?

- anonymous

You meant that proposition is not a tautology.

- anonymous

As far as I know a tautology is a predicate that is always true for any value of their parameters.

- anonymous

\[P\wedge Q\] is a predicate that can be true or false. So it's not a tautology.

- anonymous

Anyway I appreciate your help.

- anonymous

You're claiming it can be false... The point is to get you to prove it without counterexample.

- anonymous

I'm just curious as to the rules of the game.

- anonymous

I think I'm not using a counter example and that is a valid proof.

- anonymous

A tautology is a restatement within the same premiss like all trees are made of wood

- 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

I don't think saying "is a predicate that can be true or false." is valid without a counter example.

- anonymous

Or maybe there is another way, but I'm not sure

- anonymous

Is not evident that that predicate can be true or false. Its not a valid proof.

- anonymous

and \[P\wedge Q\]is by definition true or false.

- anonymous

Sure but then what makes something 'evident'?

- anonymous

Is there some standard form you want it to be in?

- anonymous

I wanted to say that there is a definition by saying "evident"

- anonymous

What about \(P\wedge Q \wedge R\)?

- 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

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

let \[P\wedge Q\Leftrightarrow S \]\[S\wedge R\]

- anonymous

I mean prove without counter example

- anonymous

so it's not a tautology.

- 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

Using your logic that \(\wedge\) doesn't give a tautology, then couldn't we say \(\neg P\vee P\) isn't a tautology?

- anonymous

ITs a Syllogism if the expanded form is the antecedent, but because its the consequent it is not tautological

- 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

q's relation to p and r

- anonymous

This is not a tautology:\[P\wedge Q\], not just the \[\wedge \] symbol.

- anonymous

Sorry I did not understand that @Edutopia

- anonymous

I'm just beginning with this logic things lol.

- 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

ahh that is right Edutopia.

- 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

But maybe that is a semantic argument? It's just my thought.

- 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

Yeah it's interesting @wio and I agree your counterexamples are a good tool. I just think they are not elegant.

- 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

I'm not sure, but I think it's valid in math.If\[x=y\]then\[x+a=y+a\]

- anonymous

That is the addition property of equality.
\(Q\) was not being appended to both sides of the implication.

- anonymous

yes, but in math its all a valid argument because you are using numbers

- 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

a=a and so forth

- anonymous

@No-data I have no idea :-). I'm so stupid.

- anonymous

Is this for linear algebra?

- 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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