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 francken
  • Posts: 3
  • Joined: Oct 15, 2018
|
#60024
I am getting quite frustrated because I seem to be grasping the logic reasoning sections for the most part but then I will get stuck on a rule where I am not correctly identifying a contrapositive to determine inferences in games. When I search the subject of contrapositives in the forum it only goes to specific games but not an overall explanation. Could someone please provide a link where I can thoroughly read on this to see where I am not connecting? Specifically speaking, I seem to be missing where I can switch terms (for example, lesson 5-104 game #1, rules 3&4, 5-106 game #2, rules3 &4, 5-118 rules 2&3, lesson 7 game #2, question #9)

Thanks
 James Finch
PowerScore Staff
  • PowerScore Staff
  • Posts: 943
  • Joined: Sep 06, 2017
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#60160
Hi Francken,

Whenever presented with conditional reasoning, you'll want to also be making note of the contrapositive, as inferences can derive from chaining either the conditionals as given, or one conditional relationship and the contrapositive of another. In addition, games that are overloaded (more variables than slots) or underfunded (more slots than variables) will often have inferences that derive from forcing certain variables into or out of a group, based on other variables being forced in or out.

To illustrate these principles, let's look at the first game you mentioned (Game 1, June 2003 Q 18-23 on page 5-104). We're given 9 variables, divided into 3 subgroups of 3 each, and only 6 slots to put them in. So first thing to notice is we'll always have exactly six in the group, so exactly three must always be out. Remember this, as it will come into play when we deal with chaining the rules together.

Our variables are:

Monkeys: F, G, H
Pandas: K, L, N
Raccoons: T, V, Z

Our rules state that:

1. F and H are not both selected. Diagrammed, this could become

F :dblline: H

However, because of the possibility of chain relationships, I prefer to diagram both possibilities as:

F :arrow: H,

and its contrapositive:

H :arrow: F

2. N and T are not both selected. Same deal as rule one, so:

N :arrow: T

and

T :arrow: N

3. If H, then K. Pretty simple, this diagrams out to:

H :arrow: K

and

K :arrow: H

4. If K then N, or:

K :arrow: N

and

N :arrow: K

Now a clear chain relationship takes shape as the last rule ties the rest of them together. So we can create two conditional chains, with one being the contrapositive of the other, along with the H/F being represented as well:

H :arrow: F (note in my actual diagram, I have this coming off the same H as below, but above onto a higher line)
H :arrow: K :arrow: N :arrow: T

and

T :arrow: N :arrow: K :arrow: H
F :arrow: H (same as the earlier chain, except this time the F is coming from below and leading into my H)

Now note that only 5 variables are affected by the rules, while 4 are not. These 4 are effectively wildcards, and can always be in or out, as needed to fill up the group. But remember, we can only ever have 3 out, which means that if we have T in the group or N out of it, that forces a group of: F G L T V Z, as N K and H all must be out. Contrast that to our contrapositive chain on top, where having H only forces out two variables, F and T, leading one slot available for either V or Z. With this setup, all but the final question (#23) become dead simple to answer, and even that one isn't too hard when you go back and look at which variables are which kind of animals.

I've attached my horribly scribbled diagram as well.
Image

Hope this helps!

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