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Formal Logic Additive Inference Drill, Q#4

LSAT Novice
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Joined: Mon Sep 24, 2018 11:32 am
Points: 1

This question pertains to the Formal Logic Additive Inference Drill, Q#4, which reads:
Some Ts are Us
All Us are Vs
All Ts are Ss

The answer key to this drill lists three additive inferences as correct:
U :some: S
T :some: V
S :some: V

The first two additive inferences make sense. However, the last additive inference (S :some: V) is a bit confusing and I wonder how we can arrive at the answer provided the 11 Principles listed in the LR Bible Formal Logic chapter. Principle #9 concerns "some" and most" combos. The principle states there will be no inference, when two "some"s are next to each other (e.g., A :some: B :some: C).

Returning to drill question #4, we can arrive at the equation V :some: U :some: S. We arrive at this equation by substituting one of the first two inferences into the original diagram. However, what we are left with is a "some" combo (i.e., two "some"s next to each other). We are supposed to derive S :some: V, according to the answer key. However, Principle #9 tells us specifically that there will be no inference if two "some"s are next to each other. How then are we supposed to derive V :some: S? Are there exceptions to the "some" "some" combo like there are for the "most" combo?

The drill answer key says "see the book website for an expanded explanation." So, I am seeking clarification.

Thanks for any assistance!
Malila Robinson
PowerScore Staff
PowerScore Staff
Posts: 296
Joined: Thu Feb 01, 2018 10:41 am
Points: 294

Hi mcharlt,
For ease of explanation let's use "at least one" for "some". So if we are given:
Some Ts are Us
All Us are Vs
All Ts are Ss
And we want to know how S :some: V
This means that:
Every T gets an S, At least one T(which also has an S) gets a U, Every U gets a V. So there is at least one TSUV (which is probably more clearly diagrammed as TS :some: UV).
Hope that helps!