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

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

@brinethery Do you have any idea?

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

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

^^ makes alot of sense

I did not say it was true

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

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.

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?

Which theorems?

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.

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

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

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

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

I mean prove without counter example

so it's not a tautology.

q's relation to p and r

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

I'm just beginning with this logic things lol.

ahh that is right Edutopia.

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

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

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

a=a and so forth

Is this for linear algebra?