Double not arrow

[Antal]

Hello again,

If you have

A B

Isn't it different from

B A ?

If you break it down you get different necessary and sufficient conditions.

For example,

Under the rule

A B

You have a rule: if not A then B

Whereas under

B A

That rule/inference doesn't exist.

So my real question is, I'm working on LGB chapter 5, how do you use those ( A B,
B A) interchangeable in the drills if they yield different inferences?

It seems as though on which side of the double not arrows the variables are placed is important.

[Nikki Siclunov]

Hi Antal,

Thanks for your question. The Double-Not arrow connects interchangeable propositions: indeed, A B is logically equivalent to B A. In both instances, A and B cannot both be selected (if A is selected, then B cannot be selected; and vice versa). The Double-Not arrow is simply a shortcut representing the following conditional relationship (and its contrapositive):

A NOT B
B NOT A (contrapositive)

In both instances, the necessary condition is negated: selecting either A or B is sufficient to ensure that the other one cannot be selected. In other words, at most one of A or B can be selected.

Now, compare the above rule to the following:

NOT A B
NOT B A (contrapositive)

The latter entails a very different outcome: if either A or B is not selected, then the other one must be selected. In other words, at least one of A or B must be selected (and possibly both can be selected). We avoid using the Double-Not arrow to represent the latter proposition, as it results in a rather confusing diagram (NOT A NOT B).

You can read more about the two conditional relationships here:

http://blog.powerscore.com/lsat/the-mos ... -rule-lsat

Hope this answers your question!

Thanks,

[Antal]

Sorry to be a pain, and I might be over thinking this but I'm having trouble understanding how

A B and B A are equivalents.

To use the example in the LGB it says that:

R H means:

R Not H
Not R H
Not R Not H

Now if you take

H R

You have:

H Not R
Not H R
Not H Not R

And when you line up up the inferences between of the two you get different sufficient and necessary conditions. This is what I meant from my previous post. See below,
Set 1____________________________________Set 2
R H___________________________H R

#1. R Not H____________________H Not R
#2. Not R H____________________Not H R
#3. Not R Not H________________Not H Not R

(Sorry about the lines btw the sets for some reason that's the only way I could keep the spaces)

What I was asking was if you look the inferences for set 2, #2 R is a necessary condition and therefore can occur without not H. But in set 1, #1 if R occurs then it must occur with not H. Do you see why I'm confused? How can R both occur with not H and without not H?

[Dave Killoran]

Hi Antal,

Thanks for the question! I think I see where you are running into problems, and it starts right here—the following is incorrect:

R H means:

R Not H
Not R H
Not R Not H

Now if you take

H R

You have:

H Not R
Not H R
Not H Not R

That's not what those mean (only the first statement under each double-not arrow diagram is correct, actually), and it's definitely not what the LGB says The " " symbol simply means that the two variables can't go together. So, if I rework the above section, it would look like this:

• R H means:
  
  R Not H
  Contrapositive: H Not R
  
  Not R ? — If R does not occur, H can occur or not occur.
  Not H ? — If H does not occur, R can occur or not occur.
  These two statements in combination mean that if you have one of R or H, then the other cannot occur. In simple terms, they cannot occur together.
  
  Basically, under R H only one scenario is outlawed:
  
  R and H both occur — this cannot occur.
  
  Three scenarios can occur:
  
  R occurs and H does not occur.
  H occurs and R does not occur.
  Neither R nor H occurs.
  
  Now if you take
  
  H R
  
  You have:
  
  H Not R
  Contrapositive: R Not H
  
  Not H ? — If H does not occur, R can occur or not occur.
  Not R ? — If R does not occur, H can occur or not occur.
  Note that these two statements are the same as above, just in reverse order, which has no effect. These two statements in combination mean that if you have one of H or R, then the other cannot occur. In simple terms, they cannot occur together.
  
  Basically, under H R only one scenario is outlawed:
  
  R and H both occur — this cannot occur.
  
  Three scenarios can occur:
  
  R occurs and H does not occur.
  H occurs and R does not occur.
  Neither R nor H occurs.

So, the two are indeed identical. Let's start there, and if you could, take a few minutes and compare what I've written to your original. After looking it over, does that help explain why the two are equivalent?

Please let me know. Thanks!

[Antal]

Thanks for the reply! I appreciate it, it does help.

My confusion must have resulted from not remembering that both Not R and Not H can both occur together. Or that if for example if Not H occurs then R can or cannot occur. That Not H doesn't necessitate R.

I have a specific question although. What would be the implications if you took the contrapositive of:

Not R H or Not H?

Thanks,
Antal Viczian

[Dave Killoran]

Hi Antal,

I'm glad that's helped so far!

Looking at the new rule you asked about, did you mean for the necessary condition to reference H for both variables? Or is this coming from you original question (which, if you recall, had some issues that lead to invalid results)? Overall, as a statement, you wouldn't see this very often (off the top of my head I don't recall anything similar), because the "or" means that either state of H could occur. In other words, anything goes. If we have a binary system where you always have H or not H, then this rule doesn't mean anything. It just means that when R isn't there, H could or could not be there (so, as far as H is concerned, anything can happen, which means the rule doesn't tell us anything new or even involve an actual restriction). I can make up cases where this would come close to being a possible rule (and would involve Grouping scenarios where there are multiple selection stages or where there are multiple Hs in play), but I'm not sure that's really useful here.

If you changed the rule so that the "or" was an "and," and you also knew this was a binary system with a single H available for selection (along the lines of a standard grouping game), what you could conclude is that "not R" would be impossible, and that therefore R would have to occur. A rough analogy would be the following:

• "To be admitted to Harvard Law, you must be rich. But I've just learned that to be admitted to Harvard Law you cannot be rich. Therefore, no one can be admitted to Harvard Law."
  
  If "rich" is used in a consistent sense (such as referring to monetary wealth in both instances, then you can't have someone who is both rich and not rich. The binary system wouldn't allow for those two opposites to jointly occur with one person. Thus, since the necessary condition is impossible, the contrapositive tells us that sufficient condition also can't occur, meaning no one is going to Harvard Law.
  
  If you change the "and" to an "or," you can see how this rule isn't that helpful:
  
  "To be admitted to Harvard Law, you can be rich but you also don't have to be rich. So, to be admitted to Harvard Law you are either rich or not rich."
  
  It's like, yeah, well so what then? Anything is still possible.

So, let me know if that rule is stated correctly, or if this relates to a specific problem you are looking at beyond our original discussion. Having more context here would help. Otherwise, you wouldn't see the rule as posed. Thanks!

[Antal]

Thanks again,

Your explanation did help.

I'm pursuing this as a hypothetical situation just to see how far I can push my logic. Could you expand on what you mean by "binary system?" I guess what I was trying to get at was what you nailed which is that "the binary system wouldn't allow for those two opposites to jointly occur with one person."

Antal

[Dave Killoran]

Hi Antal,

Binary in this situation just means two possible outcomes: it occurs or it doesn't occur, and it's always one or the other. Those two states can't co-exist, and that's something that important for that discussion to make sense (in part because that is typically how the LSAT works).

I'm a strong advocate of considering hypothetical situations—it's a really useful process! What's happened here though is that we've ventured into an area that wouldn't occur within LSAT LG (or if it did, it would require a really specific scenario where there would be a lot more info surrounding such a rule). That's why you saw me ask several setup questions to try to figure out where you were going, and to help bring that back to what we would be more likely see on an LSAT (and that's why I changed the discussion to necessary conditions that were opposite, because that can happen on the test).

Thanks!

[ryanshort8883]

Hello,

I realize this is a very old post, however I wanted to post a reply as I also ran into some confusion from the way it is seemingly presented in the book.

First, I love the PS Bibles and think they're absolutely great. I'm not trying to be overly critical. I'm a computer engineering major and as such have worked extensively with digital logic and Boolean algebra, so I thought I would attempt to clear something up.

Antal's confusion, and perhaps my confusion at first as well, came from the fact that the book (although not explicitly stated, so it could just be that I was reading into it) seems to indicate that the double not arrow is a negative biconditional. This is incorrect.

The biconditional, in digital logic, is logically equivalent to and referred to as an XNOR gate/boolean operator. Which means that a negative biconditional is an XOR gate, which only allows for the possibility of two scenarios occurring.

The double not arrow, in digital logic, is a NAND gate. This, imo, is more clear and is how I prefer to think of it. But I recognize that many prepping for the LSAT have no experience with digital logic.

Anyways, just thought I would mention this.

Thanks for all you do in helping us prepare for the LSAT!

Ryan

[Brook Miscoski]

Ryan,

Thanks for your comment. You're correct that the book does not call the scenario you described a negative biconditional, and I believe you're correct about the remainder. Our techniques are designed to be used by anyone regardless of math, engineering, or philosophy backgrounds. Sometimes that means that the language is a bit different.