Law School Discussion

question?

question?
« on: July 20, 2007, 05:00:31 PM »
All Xs are Zs All Ys are Zs

So
X-->Z<--Y
we can still conclude that some Xs are Ys..or am i nuts?  :-\
my instructor said you can't make any inferences.

Re: question?
« Reply #1 on: July 20, 2007, 05:06:12 PM »
All Xs are Zs All Ys are Zs

So
X-->Z<--Y
we can still conclude that some Xs are Ys..or am i nuts?  :-\
my instructor said you can't make any inferences.

You can't make any valid transitive inference from that set.

X --> Z

Y --> Z

Two different sufficient conditions that both lead to the same necessary condition does not guarantee any cross over amongst the two different sufficient conditions.



Oh! thank you so much!

papercranes

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Re: question?
« Reply #2 on: July 20, 2007, 05:08:41 PM »
One thing that helped me think about this was to add a little to the diagram when first working on it.

X--->Z

so if you have a bunch of Xs floating around
x    X
  X     x  x

They're actually all Zs  
xz    Xz
  Xz     xz  xz

But there's no reason that you can't have just Zs
xz    Xz    z   z  z
  Xz     xz  xz

So
X--->Z

xz    Xz    z   z  z
  Xz     xz  xz

and

Y--->Z

So you could always have:
xz   YZ Xz    z   z  z
  Xz     XYZ  xz
YZ   YZ

Or something else, but you can't know anything but what they give you.

 This does get easier without all my Zs Ys and Xs

Re: question?
« Reply #3 on: July 20, 2007, 06:26:05 PM »
One thing that helped me think about this was to add a little to the diagram when first working on it.

X--->Z

so if you have a bunch of Xs floating around
x    X
  X     x  x

They're actually all Zs  
xz    Xz
  Xz     xz  xz

But there's no reason that you can't have just Zs
xz    Xz    z   z  z
  Xz     xz  xz

So
X--->Z

xz    Xz    z   z  z
  Xz     xz  xz

and

Y--->Z

So you could always have:
xz   YZ Xz    z   z  z
  Xz     XYZ  xz
YZ   YZ

Or something else, but you can't know anything but what they give you.

 This does get easier without all my Zs Ys and Xs


Lolll thanks for the help  :)

Re: question?
« Reply #4 on: July 20, 2007, 06:29:31 PM »
Beating a dead horse, but I really liked this hypo:

All men are people.  All women are people.

Does this mean some men are women?

that's a good one actually! Merci!

Re: question?
« Reply #5 on: July 23, 2007, 11:10:32 AM »
Although I wouldn't recommend it for every question (or even most), for questions like these try a Venn diagram if you aren't sure what the set is saying. Imagine a big circle labeled Z with Z's in it, and inside the big circle of Z's are respective circles for X and Y with all the X's and all the Y's in them--only the circles of X's and Y's are a 10th the size of the Z circle and floating around in the big Z circle. Maybe the X circle is a 4th the size of Z and the Y circle is a 3rd the size. But the set does not say anything about overlap, and Venn diagrams are a good way of illustrating that X, Y, and Z do NOT have to be of equal number, even if all of one equals another. The set does say that all X's and Y's are Z's, but not how many there are of each. As another poster mentioned, you could have a billion Z's, two Y's, and one Z. Now, the X and Y circles could overlap, they could not. They could just be floating around within the bigger Z circle on opposite sides. The X circle could be as big as the Z circle, but the premise does not state that.

It's different from saying that All X ---> Y, All Y ---> Z, therefore all X ---> Z. In that case there would be three overlapping circles of the same size using Venn diagramming. You could reduce the all to some, and the Venn diagram would still illustrate that.