Diagram

[ecmoloney]

Looking for a nice clean way to draw the arrows for the "If math is offered, then either literature or soc. (but not both) is offered". Or does it just mean writing MLS crossed out on the side?

[ManhattanPrepLSAT1]

Unfortunately, there's not as simple a way as I would like. I've attached a setup of the game using the Logic Chain. The constrain cannot be built into the setup, so it is placed below as a reference.

I hope this helps!

[ecmoloney]

Thanks. I think the simple breaking out of the M is a good idea (I was keeping it in but then couldn't figure out a non-confusing way of saying if M is out...).

[gethornburg]

A game of chained inference ...

g h l m p s z

1. m > l / m > s & ~[l&s]

2. l > g

2. l > ~p

3. s > p ~p > ~s

3. s > ~z z > ~s

4. g > h

4. g > z

chains,

l > ~p > ~s
l > g > z > ~s
l > g > h

[ if l then g, h, z and ~s]

contrapositives

s > p > ~l

s > ~ z > ~g > ~l

~h > ~g > ~l

m combinations

m > l > g,h,z and ~p > ~s

m > s > p and m > s > ~ z > ~g > ~l

m contrapositives

~l & ~s > ~m


13. A

if l then g, h, z: eliminates b,d,e
m > l/s eliminatees c

14. C

if l then g, h, z and ~s: eliminates, a, b, d, e

15. A by contrapositive ~h > ~g > ~l

16. D at least 3 by m combinations

m > l > g,h,z
m > s > p

17. E by g > z > ~s

or

m combinations eliminate b, c & d
no rule preventing a

[solade1]

I noticed that the 1st rule (if M is offered then eithier L or S but not both is offered) cannot be diagramed into the logic chain, the contrapositive of this rule is If L or S is not offered then M is not offered. This contra has a "or" in the sufficient so we should be able to seperate into 2 seperate rules, however when I put these seperate rules on my logic chain it causes me to get problem #15 and#17 incorrect. Can anyone explain why my logic does not hold.

[timmydoeslsat]

I noticed that the 1st rule (if M is offered then eithier L or S but not both is offered) cannot be diagramed into the logic chain, the contrapositive of this rule is If L or S is not offered then M is not offered. This contra has a "or" in the sufficient so we should be able to seperate into 2 seperate rules, however when I put these seperate rules on my logic chain it causes me to get problem #15 and#17 incorrect. Can anyone explain why my logic does not hold.

The original conditional statement is this:

M ---> 1 L/S in, 1 L/S out

The contrapositive of this is:

Both LS in ---> ~M

Both LS out ---> ~M

The "or" statement is also a "but not both" statement, which is important logically.

So it is not true that lacking L or lacking S will result in M being out. It would take both in or both out for M to be out.

For M to be in, it must be the case that the L and the S are separated.