December 2014 LSAT
Section 2
Question 23

December 2014 LSAT
Section 2
Question 23

Reply

Irina on September 14, 2019

@Jwebb,I am not able to see the video explanation for this question, but choices (A) and (C) can be diagrammed as follows:

(A) If K is assigned to S university ceremony, then H & M cannot both be assigned to that ceremony.

K -> ~(H & M)

which is equivalent to:

K -> ~ H v ~ M

The statement tells us that if K is assigned, it must be true that either H or M is NOT because they cannot BOTH be assigned.

it is also equivalent to its contrapositive:

H & M -> ~K

If BOTH H& M are assigned, we can infer that K is not assigned. But we could not infer that if we only knew that H or M is assigned as K only precludes H&M from being assigned together, not either of them from being assigned individually.

(C) Unless K is assigned to T ceremony, both F & M must be assigned to that ceremony.

"Unless" generally introduces a necessary condition, whereas a sufficient condition must be negated.

If BOTH F & M are not assigned, K must be assigned.

~(F & M) -> K

or

If EITHER F OR M are not assigned, K must be assigned as IF EITHER is not assigned, it is impossible for BOTH to be assigned:

~ F v ~ M -> K

This statement is also equivalent to its contrapositive:

~K -> F & M

If K is NOT assigned, BOTH F & M must be assigned.

Both of the statements in (A) and (C) rely on the same rule on inference, known as De Morgan's law, that tells us that these statements are logical equivalents:

~(p & q):: ~ p v ~q

It is sometimes challenging to make inferences using abstract variables instead of natural language as is often required in logic games, so memorizing this rule of inference could be helpful in these cases.

Let me know if you have any further questions.

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