## Tough Conditional- PT 59 S2.19 Lawyers/Bankers/Athletes

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NightmanCometh

Posts: 99
Joined: Thu Jul 21, 2011 9:03 pm

### Tough Conditional- PT 59 S2.19 Lawyers/Bankers/Athletes

Came across this problem that stumped me with its conditionals (even after consulting Powerscore LR Bible section).

Premises:

Words: All of the B's are A's.
Conditional: B -> A (A -> B)

Words: None of the L's are B's.
Conditional: L -> B (B -> L)

Why is it (C) and not (B)? I can see how both of them work.

(B): Some L's are not A's.
L <-SOME-> A
Using contrapositive of premise 1, L <-SOME-> A -> B, which means L <-SOME-> B. Since no L is B (premise 1), some L is not B. So this is true, right?

(C): Some A's are not L's.
A <-SOME-> L
Using premise 1, B -> A <-SOME-> L, which means B <-SOME-> L. This is true, because since no B is L (see contrapositive of premise 2), some B is not L.

Don't see the difference between the logical conclusions reached by both answer choices...searched through a couple forums and didn't see anything. Anybody have a good answer for this?

Thanks!

suspicious android

Posts: 919
Joined: Tue Feb 09, 2010 4:54 pm

### Re: Tough Conditional- PT 59 S2.19 Lawyers/Bankers/Athletes

P: B --> A
P: L --> ~B (B -->~L)

Based on these two premises, we know two things about bankers. They are athletes, and they are not lawyers.

However, we don't really know much about lawyers or athletes as a group. The lawyers could all be athletes, in fact it could be that every member of the gathering is an athlete.

From the pattern:

A --> B (like all bankers are athletes)
A --> C (like all bankers are non-lawyers)

You can only conclude that some B's are C's (assuming there is at least one A). That's how we get "some athletes are not lawyers".

NightmanCometh

Posts: 99
Joined: Thu Jul 21, 2011 9:03 pm

### Re: Tough Conditional- PT 59 S2.19 Lawyers/Bankers/Athletes

Thanks, makes sense!

But, the thing that trips me up is answer choice (B)...did I do something wrong with the conditionals there in my chain of logic? Because following the conditional rules, the logical conclusion that is reached is also true.

(B): Some L's are not A's.
L <-SOME-> A
Using contrapositive of premise 1, L <-SOME-> A -> B, which means L <-SOME-> B. Since no L is B (premise 1), some L is not B.

I know that (C) is correct, just would like to know if possible why (B) is NOT correct...

jamesireland

Posts: 43
Joined: Sun Dec 05, 2010 8:52 pm

### Re: Tough Conditional- PT 59 S2.19 Lawyers/Bankers/Athletes

The only way I see to join to the two premises is: ~L <- B -> A (~A -> ~B <- L).

'B' must link them because that is all they have in common.

To get that some L's are not A's, you would need something like this: L <-S-> ~B -> ~A (The arrow needs to lead away from the 'some'). The contrapositive above has the arrow from '~B' to '~A' pointing in the wrong direction to make this sort of inference.

suspicious android

Posts: 919
Joined: Tue Feb 09, 2010 4:54 pm

### Re: Tough Conditional- PT 59 S2.19 Lawyers/Bankers/Athletes

[quote="NightmanCometh"]

(B): Some L's are not A's.
L <-SOME-> A

[quote]

This is the error. Some L's are not A's does NOT mean that some L's are A's.