A Few More Words about Qualifiers

Posted on January 21, 2013 by Anthony Ritz

A few months ago I wrote some words about qualifiers. These little words — words like “some,” “many,” “most,” “few,” “maybe,” and “probably” — somehow manage to be both often-critical and often-overlooked on standardized tests like the LSAT, the GMAT, and the GRE. I discussed spotting these words and went on to point out the existence and importance of hierarchies among the various qualifiers.

There’s one more twist, though, that we have yet to address…

Consider the statement “Some $A$ s are $B$ s.” Simple, straightforward… but what if you had to find a different, equivalent version of this statement? What if, for instance, I asked — nay, demanded — that you draw a logical inference from the following set of arrow diagrams?

$\begin{tabular}{ c} some PC \rightarrow FI\\ PC \rightarrow !LS\\ some PC \rightarrow WS\\ \hline \\ \end{tabular}$

It’s a problem, right? They don’t seem to fit. The only common element among the diagrams is $PC$ , but $PC$ is at the beginning of each diagram and never at the end. The transitive property, you’ll recall, only works if the end of one arrow diagram exactly matches the start of another.

The contrapositive won’t help either, for two reasons. First, taking the contrapositive of any of these arrow diagrams will introduce a “not” to each side of the flipped diagram. So it seems like $PC$ might end up on the right side, but it will actually be $!PC$ , and it still won’t match or link with the $PC$ s at the beginnings of the other diagrams. Second, and even more fundamentally, it is impossible to take the contrapositive of an arrow diagram with a qualifier. Why? Because an arrow diagram with a qualifier after its arrow simply makes no sense: “Things that are not $FI$ s are not some $PC$ s”? What does that even mean?

So neither the transitive property nor the contrapositive will save us here. Some amount of clever maneuvering is clearly required.

To take the next step, let’s pause to consider what the statement “ $some$ $A \rightarrow B$ ” actually looks like. Here we have circles representing “things that are $A$ s” and “things that are $B$ s”:

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Since “some $A$ s are $B$ s,” these circles must overlap. And where are these “some $A$ s” that are $B$ s? Well, in the overlap, of course!

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What else can we conclude from this picture? Well, what if I claimed that “some $B$ s are $A$ s”? That has to be true as well, doesn’t it? After all, we can point to the exact same shaded region. So it should be clear that if “some $A$ s are $B$ s” then “some $B$ s are $A$ s” and — put yet a third way — that “there exist items are are both $A$ s and $B$ s.”

In formal logical terms, we can write $some$ $A \rightarrow B = some$ $B \rightarrow A = A + B$ .

Now let’s see that inference question again:

$\begin{tabular}{ c} some PC \rightarrow FI\\ PC \rightarrow !LS\\ some PC \rightarrow WS\\ \hline \\ \end{tabular}$

Take a minute and try your hand at working out the deductions.

…

Using the new rule $some$ $A \rightarrow B = some$ $B \rightarrow A$ , we can transform the first and third arrow diagrams as follows:

$\begin{tabular}{ c} some FI \rightarrow PC\\ PC \rightarrow !LS\\ some WS \rightarrow PC\\ \hline \\ \end{tabular}$

At last we’re able to apply the transitive property! Connecting the first and second diagrams gives:

$some$ $FI \rightarrow !LS$

Connecting the second and third gives:

$some$ $WS \rightarrow !LS$

Of course each of these statements can be flipped by the “some” rule as well:

$some$ $!LS \rightarrow FI$

$some$ $!LS \rightarrow WS$

Note that no amount of manipulation will make the first and third rules connect. Flipping just the first, for instance, would produce:

$\begin{tabular}{ c} some FI \rightarrow PC\\ some PC \rightarrow WS\\ \hline \\ \end{tabular}$

However, as we saw in Formal Logic 102, the “some” at the start of the second statement is a monkey wrench that ensures that the two diagrams cannot connect.

Now, lest you think that this is all just pedantic navel-gazing, let’s take a look at these issues on an actual test question. How about LSAT PrepTest 29, Section 1, Question 18 (Next 10 LSAT, page 21).

This is a formal logic-based question, so take a minute to try your hand at arrow diagramming the premises.

…

The first statement is comparatively straightforward:

$some$ $PC \rightarrow FI$

The second is far trickier. Simply reading from left to right gives $!PC \rightarrow LS$ , but this isn’t correct. Don’t believe me? Read it back: “If not PC, then LS.” That’s definitely not what the argument says, though.

As discussed in Formal Logic 102, the trick is to first paraphrase into an if-then statement and only then translate the paraphrased statement into formal logic. And don’t forget to confirm your arrow diagram by reading it back at the end and comparing it to the original statement. “If this person is a planning committee member, then this person does not live in the suburbs”:

$PC \rightarrow !LS$

The third statement isn’t as tricky, but it’s easy to miss since it’s tacked on as the last clause of the same sentence — “many of them work there.” To make life a bit easier, let’s treat that “many” as “some.” The principle of weakening establishes that if “many” have a certain characteristic, then surely at least “some” do.

$some$ $PC \rightarrow WS$

All together now:

$\begin{tabular}{ c} some PC \rightarrow FI\\ PC \rightarrow !LS\\ some PC \rightarrow WS\\ \hline \\ \end{tabular}$

Hey… wait a minute! That looks too familiar. Isn’t that the same set of arrows we examined back in the pedantic navel-gazing section? My gosh, it is! Well in that case, we already know the possible inferences:

Connecting the first and second diagrams gives:

$some$ $FI \rightarrow !LS$

Connecting the second and third diagrams gives:

$some$ $WS \rightarrow !LS$

It’s just a matter of finding one of these options in an answer choice. A quick scan reveals that we’re in luck. Answer E says $some$ $FI \rightarrow !LS$ . So E is correct.

The rule that $some$ $A \rightarrow B = some$ $B \rightarrow A$ can be quite useful, but it’s important not to push it beyond its limits. In our previous look at qualifiers, we noted important differences among qualifiers of different strengths. Here, too, different qualifiers play by slightly different rules. In fact, it turns out that the rule we’ve learned today will work only with the qualifier “some.” That is:

$many/most/few$ $A \rightarrow B \not= many/most/few$ $B \rightarrow A$

Here we have a diagram in which most $A$ s are $B$ s, and most $B$ s are $A$ s as well.

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But in this diagram, although most $A$ s are still $B$ s, the reverse is no longer true — most $B$ s are in fact not $A$ s.

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So for any qualifier other than “some,” the best we can do is to weaken our proposition to a “some” statement and then flip that. After all, in the above diagrams, we can still see that at least some $B$ s are $A$ s in any case.

While we’re pointing out differences among qualifiers, here’s one more. Since “most” means “more than half,” two subsets that each encompass “most” of a set cannot help but overlap. That is,

$\begin{tabular}{ c} most A \rightarrow B\\ most A \rightarrow C\\ \hline some B \rightarrow C\\ \end{tabular}$

These differences between “some” and other qualifiers such as “many” and “most” are tested on these exams as well. For example, look at LSAT PrepTest 30, Section 2, Question 17 (Next 10 LSAT, page 59).

Both the phrase “a very small percentage” from the first sentence (the premise) and the word “underrepresented” in the second sentence (the conclusion) essentially boil down to meaning “few.” See if you can diagram out the propositions here and come up with the right answer.

…

Using “few” in each proposition, as suggested above, the first statement, the premise, can be written as:

$few$ $SP \rightarrow BM$

The second statement is the conclusion — note the phrase “this shows that.” Diagramming it is tougher, but paraphrasing can help. When we say that $SP$ s are underrepresented among $BM$ s, what we really mean is that “few $BM$ s are $SP$ s”:

$few$ $BM \rightarrow SP$

Taken together, the argument structure is:

$\begin{tabular}{ c} few SP \rightarrow BM\\ \hline few BM \rightarrow SP\\ \end{tabular}$

It seems that the argument has swapped the order of its “few” statement between the premise and the conclusion. That works great with “some,” but now we know that you just can’t do that with any other qualifier. Even if few members of service professions are board members (and, in fact, few of anybody are board members), many, most, or even all board members could be members of service professions. This premise just isn’t very useful for establishing the conclusion.

Running through the answer choices:

Answer A is incorrect. The problem here isn’t one of sample size; the boards of “the 600 largest North American corporations” cover most if not all of “the most important corporate boardrooms in North America.” Regardless, no amount of increase of the data set will fix the problem discussed above.

Answer B is correct; as previously noted, “few $SP$ s are $BM$ s” just doesn’t say much about the claim that “few $BM$ s are $SP$ s.”

Answer C accuses the argument of an assumption it never made. The conclusion of the argument limits itself to “the most important corporate boardrooms” and never purports to speak to “corporate boardrooms generally.”

Answer D is dead on arrival as soon as it mentions the entirely-irrelevant “smaller corporations.” Again, the conclusion only ever claims to speak to “the most important corporate boardrooms.” Failing to draw a conclusion that some parties might find “relevant” is not a flaw.

Answer E heads even further out to left field. The argument never makes any claim either way about what sorts of corporate boards best promote social responsibility. Don’t impose your own values, expectations, preconceptions, or speculations onto the argument. Your only job is to determine whether the given conclusion follows from the given premise(s) and, in this case, to explain why the argument falls short of that singular goal.

The “some” rule, $some$ $A \rightarrow B = some$ $B \rightarrow A = A + B$ , adds yet another weapon to our arsenal for attacking the LSAT and other tests involving logical reasoning. It’s a valuable piece, if used correctly.

Look out for this trick whenever a question features out multiple “some” statements — especially if formal logic seems to be involved. Just be careful not to wrongly apply this trick to statements involving stronger qualifiers that won’t hold up when reversed.

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