Logic is confusing can anyone explain this?

>>8972362
It is true if P and Q are both true.
It's also true if P is false

When you read "if P then Q" you should interpret it as "if P is true then Q must be true".

It says nothing about what happens if P is false. If P is false there is no requirement.

>>8972362
The case where P(x) is false and the statement is true is an example of vacuous truth. I like to think of it this way:
I tell you "If you give me 20 dollars, I'll give you an A in the class", if you don't give me 20 dollars and I give you a B, did I lie? In logic, the answer is no.

>>8972362
>>8972371
Also with the universal, if you want the universal to be true [math]\forall x(P(x) \rightarrow Q(x))[/math], this universal is saying "All things that hold the property P, also hold the property Q" it says nothing about things that are Q and not P, nor does it say anything about things that are not P.
[math]\forall x(P(x) \wedge Q(x))[/math]
Says that everything is both P and Q, which is a much stronger statement.

>>8972368
>>8972371
Thanks.

>When you read "if P then Q" you should interpret it as "if P is true then Q must be true".


So, I should only consider the cases where for all x in the Domain P(x) is true.... in other words only consider when P(x) is true?

I'm just trying to wrap my head around what the book is trying to explain..

If I understand you correctly, then I should just look at it like "When P(x) is true, then...."

Thanks this is helping a lot.

>>8972386
The statement is just saying "All x in P is also in Q" Not necessarily "All x in Q is also in P"
The interesting cases are when P(x) is true, but the statement still holds for non-P things.

>>8972395
Okay thanks. I'll think of it as "Only consider it when x in P are true, then ensure Q also holds"

In other words T->*, where * depends on Q

This belongs in /sqt/

Also read first couple chapters of How To Prove It

>>8972404
basically if P is true, then Q should also be true

and not caring what happens if P is false.

So, only the case when P is true, do I care what Q is

>>8972405
Thanks, I just got the book. Next time I will post in sqt

>>8972386
I'm the guy you replied to.

> I should only consider the cases where for all x in the Domain P(x) is true.... in other words only consider when P(x) is true

not quite, but something similar.

You might as well only consider the values of x such that P(x) is true. It makes no difference what happens when P(x) is false.

What you need to check is this: for every value of x such that P(x) is true, make sure that Q(x) is also true.

For the values of x where P(x) is false, it doesn't matter whether Q(x) is false or true.

The only thing that matters is that whenever P(x) is true, Q(x) is also true. That's precisely what it means when we say
[eqn]\forall x. P(x) \Rightarrow Q(x)[/eqn]

>>8972424
Excellent, thanks.

>>8972362
I promised my kids if they win, I buy them pizza. If I buy them pizza anyway if they lose, did I break my promise?

>if both P(x) and Q(x) are true
So P(x) if and only if Q(x)

[math]\forall x (Px \rightarrow Qx)[/math] has the same truth conditions as [math]\lnot \exists (Px \and \lnot Qx)[/math].
It does not have the same truth conditions as [math]Px \rightarrow Qx[/math], which does not have any truth conditions since it is not a well-formed formula (because the variable ([math]x[/math]) is unbound by any quantifier).

>>8973939
didn't know [math]\and[/math] didn't work
guess it's [math]\land[/math] then?

>>8972790
>>if both P(x) and Q(x) are true
>So P(x) if and only if Q(x)
No, what he's talking about is not an equivalence, it's a logical and