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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!

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".

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...

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.

[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.