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