Logic Help?

[Mr 3vil]

I'm taking a Intro course to logic, but I have no way to derive the solution from Q. By the way, we haven't got to subproofs yet.

Here's the problem.

(P disjunct of Q) biconditional P Therefore Q then P

In symbols...

(PvQ) <==> P l- Q => P

Can someone explain what I should do?

[Jinjonator]

I'd love to help you, but I haven't a clue what any of that means.

[Theguyoverthere]

So... P disjuncts Q if and only if P, and therefore Q then P?

...Wait, what?

[Avarice]

42?

Seriously, I have been 'd.

[ConkerKing]

Mr. Green with the candle in the Billiard Room.

[Jinjonator]

Originally Posted by ConkerKing

Mr. Green with the candle in the Billiard Room.

We have a winner.

[Avarice]

Originally Posted by Jinjonator

We have a winner.

What was your first clue?

Ha ha ha ha ha ha ha...

ON-TOPIC: I still have no idea...

[wkw427]

If P=Q^x4, then y=p^-8/Q...

L0+lWT/F=42

lol wtf 42..

[firealpha]

It's Greek to me.

Maybe you could say it in English?

[repobanjo]

Originally Posted by Mr 3vil

I'm taking a Intro course to logic, but I have no way to derive the solution from Q. By the way, we haven't got to subproofs yet.

Here's the problem.

(P disjunct of Q) biconditional P Therefore Q then P

In symbols...

(PvQ) <==> P l- Q => P

Most.......illogical, sir.

[Grey Jinjo]

I'm not sure, but I'll take a crack at it (a very wrong crack at it. I haven't really had a logic course, yet). I have no idea what the l- is, though, so I have some "fill in the blank" sections . Also, different professors make you write it in different ways, despite the fact that there is a global standard. This might not be as technical as you have to do it.

Note: I'm not sure if this is even what you are asking for .

1. For (PvQ) <==> P l- Q => P to be False, the left side (PvQ) would have to be true while the right side (P l- Q => P) has to be false.

2. For (P l- Q => P) to be false, P l- Q has to be true while P is false.

3. If P is false, then to make P l- Q true Q has to be false.

4. But if P is false, and Q is false then (PvQ) can never be true. Thus, the first statement of the biconditional can never be true while the second statement is false, so the biconditional is always true.

Edit: Looking back, I might have screwed up with the "if and only if" bit. I didn't notice that there were two different arrow types. Also, I have no idea what l- is suppose to represent like I said, so.

[Gaming Master2k]

I did logic last semester (and another time in 2006). When I get home from work I'll grave my textbook and work it out for you

Hopefully you don't need it sooner

[Mr 3vil]

Originally Posted by Gaming Master2k

I did logic last semester (and another time in 2006). When I get home from work I'll grave my textbook and work it out for you

Hopefully you don't need it sooner

Alright, that would be awesome, any help is appreciated. I hear it gets easier after a while.

The l- symbol means "therefore." It is supposed to be a sideways "T."

[Gaming Master2k]

I solve questions like these with input/output tables. I.e. if you have an input of P = False and Q = False you get... etc. That's how I was taught so I'm hoping you have been as well.

Ok so here we go (I'm going to go into more detail than will probably be needed so that if others want to follow along they want):

P and Q can only be True (1) or False (0).

This means we have the table

P|Q
-|-
0|0
0|1
1|0
1|1

The first part says (PvQ) which is if P is true or Q is true then PvQ is true. Therefore for the above inputs we get:

P|Q|(PvQ)
-|-|-------
0|0| 0
0|1| 1
1|0| 1
1|1| 1

If we use A and B as inputs then A <==> B (A if and only if B) has an output table that looks like:

A|B| A <==> B
-|-|----------
0|0| 1
0|1| 0
1|0| 0
1|1| 1

We can substitute A for (PvQ) and B for P. If we work it out we get:

P|Q|(PvQ) |(PvQ) <==> P
-|-|----------------------
0|0| 0 | 1
0|1| 1 | 0
1|0| 1 | 1
1|1| 1 | 1

This is the left side of the equation completed.

If we use C and D as inputs then C => D (if C then D) has an output table of:

C|D| C => D
-|-|--------
0|0| 1
0|1| 1
1|0| 0
1|1| 1

Here we substitute Q for C and P for D giving us an output table (for the Right hand side) of:

P|Q|Q => P
-|-|-------
0|0| 1
0|1| 0
1|0| 1
1|1| 1

Clearly the result of the left hand side of the equation is equal to the right hand side of the equation so:

(PvQ) <==> P is therefore (or equal to) Q => P

I hope that has cleared things up for you, or at the very least gotten you on the right track, but that's how I would have answered a question like that last semester (when I did the subject).

Feel free to ask me anything else if you don't understand where I've done something.

[Articerile]

Erm, GM? I thought PvQ would only be true if it was OR not AND/OR

P|Q|(PvQ)
-|-|-------
0|0| 0
0|1| 1
1|0| 1
1|1| 0

Where either P or Q is 1, but not both.