# Is this sample LSAT question / answer based in fallacy?

Here is a sample question from the Law School Admission Council's website:

A university library budget committee must reduce exactly five of eight areas of expenditure—G, L, M, N, P, R, S, and W—in accordance with the following conditions:

If both G and S are reduced, W is also reduced.
If N is reduced, neither R nor S is reduced.
If P is reduced, L is not reduced.
Of the three areas L, M, and R, exactly two are reduced.

Question 1

If both M and R are reduced, which one of the following is a pair of areas neither of which could be reduced?

A. G, L
B. G, N
C. L, N
D. L, P
E. P, S

They reason the answer is C

This question concerns a committee’s decision about which five of eight areas of expenditure to reduce. The question requires you to suppose that M and R are among the areas that are to be reduced, and then to determine which pair of areas could not also be among the five areas that are reduced.

The fourth condition given in the passage on which this question is based requires that exactly two of M, R, and L are reduced. Since the question asks us to suppose that both M and R are reduced, we know that L must not be reduced:

Reduced: M, R
Not reduced: L
The second condition requires that if N is reduced, neither R nor S is reduced. So N and R cannot both be reduced. Here, since R is reduced, we know that N cannot be. Thus, adding this to what we’ve determined so far, we know that L and N are a pair of areas that cannot both be reduced if both M and R are reduced:

Reduced: M, R
Not reduced: L, N
Answer choice (C) is therefore the correct answer, and you are done.

My issue being with the "... if N is reduced, neither R nor S is reduced. So N and R cannot both be reduced." It strikes me as common sense wrong, based on my understanding of logic wrong, and I would think is a well known form of misunderstanding of modes ponens. Am I missing something in the language here?

• [If R is reduced] (simplified Q1 condition), N can’t be reduced because [N being reduced makes R not reduced] (simplified condition 2). R can’t be reduced and not reduced. Maybe seeing it in this order will make the language clearer. To be clear the order doesn’t change the underlying logic. I might suggest Velleman’s How to Prove It for proving A->B proofs for more general comfort of these types of problems. There are sections for how to approach most common types of proofs (proofs have a logical backbone, it’s not just for math). You don’t need to read the whole thing. Commented Jul 25 at 22:21

It may be easier to represent this with propositional logic. Let each letter X represent the proposition "X is reduced." So we have propositions G L M N P R S W

``````If both G and S are reduced, W is also reduced.
If N is reduced, neither R nor S is reduced.
If P is reduced, L is not reduced.
Of the three areas L, M, and R, exactly two are reduced.

... both M and R are reduced
``````

1. GS -> W
2. N -> ~R ^ ~S
3. P -> ~L
4. Exactly two of L, M, R
5. MR

From 5 and 4 we conclude ~L

From 5 and 2 we conclude ~N (because we know R, so ~R ^ ~S must be false, so N must be false by modus tollens).

As an alternative explanation for why we conclude ~N, a conditional statement is logically equivalent to its contrapositive, and the contrapositive of "N -> (~R and ~S)" is "(R or S) -> ~N", and here we have R, implying ~N.

• @Spencer We know that R (premise 5). Therefore, the premise 2 consequent ~R ^ ~S must be false, therefore the antecedent N must be false. That's modus tollens. Commented Jul 26 at 15:44
• So, let me be realistic then. A = "it is morning" B = "their are birds chirping" Let's just say reality, where I live, is such that every single morning birds are chirping. So, A -> B. B does not imply A. Birds can be chirping in the evening, at night, any time. Commented Jul 26 at 15:48
• @Spencer But that's not what's happening here. You said in your first comment that "we know that ~R." We don't. We know R, not ~R. Affirming the consequent is a fallacy, denying the consequent is modus tollens, and valid. Commented Jul 26 at 15:50
• Apologies, misspoke. Hmm. Commented Jul 26 at 16:06
• I understand now. Apologies. I've been out of academia too long. I augmented the answer to assist with other people sharing my confusion. Commented Jul 26 at 16:43

The answer C is correct, but the wording of the explanation is clumsy.

The given condition, "If N is reduced, neither R nor S is reduced," tells us that if N is reduced, R is not and also S is not. We can think of this as two conditionals: if N is reduced then R is not, and also, if N is reduced then S is not. So, since we are given the assumption that R is reduced, we can infer N is not by modus tollens.

Unfortunately, the given explanation glosses the condition as, "So N and R cannot both be reduced". This is poor wording, because it would naturally be read to mean, "it is not the case that both N and R are reduced" whereas it is better expressed as, "both N and R are not reduced". This doesn't change the fact that answer C is correct.

• @Spencer R ist reduced by setup of the question. That means N is not reduced. Because if N were also reduced, condition 2 would be violated (N => not R). The implication A=>B always entails the implication not B=>not A. Commented Jul 26 at 10:16
• I disagree with the last paragraph of this answer, at least insofar as I can understand it. I agree that "So N and R cannot both be reduced" actually means "it is not the case that both N and R are reduced;" both statements are natural (albeit complex) and convey the intended meaning. To me, "both N and R are not reduced" is not a better expression; it is somewhat unnatural English, and rather ambiguous. It sounds like it could be intended to mean "neither N nor R are reduced," instead of "it is not the case that both N and R are reduced." Commented Jul 26 at 16:24
• I'm afraid I don't follow. First, the question doesn't state "not-N and not-R". I don't see where that comes from. It does say [A] "If N then (not-R and not-S)." Second, yes the explanation does say/imply [B] "not-(N and R)", but that isn't misleading; it's a correct implication of [A]. I'm wondering if your objection is that [A] and [B] are not equivalent. This is true, they aren't, but the answer isn't presenting them as equivalent. It's saying that [A] implies [B], which is true. Maybe I'm missing something ... ? Commented Jul 26 at 21:17
• @LarsH Yes, I got the letters a bit mixed up there, I should have said, "If N is reduced, neither R nor S is reduced," means, "if N then (not-R and not-S)". This is not the same as, "if N then not-(R and S)". The latter does not allow us to infer not-N from R. Commented Jul 26 at 21:56
• @Spencer "A -> B, and B being true, does not imply A is true." You misread. Here B is not true, hence A is also not true. Indeed, if B were true, we could not infer anything additional about A. Commented Jul 27 at 0:21

My issue being with the "... if N is reduced, neither R nor S is reduced. So N and R cannot both be reduced."

I'm not sure why you have an issue with this.

First, I'm pretty sure you would agree that "neither R nor S is reduced" entails "R is not reduced", right? (This is just conjunction elimination.)

Applying this conjunction elimination to the consequent of the implication "if N is reduced, neither R nor S is reduced" then entails "if N is reduced, R is not reduced". I assume you still agree with this, right?

Now, I hope I can convince you that all of the following statements are logically equivalent:

1. "if N is reduced, R is not reduced"
2. "if R is reduced, N is not reduced" (contrapositive of 1)
3. "either N or R is not reduced"
4. "N and R are not both reduced" (application of De Morgan's law to 3)

In particular, letting A stand for the statement "N is reduced" and B for "R is reduced", the four statements above can be written using conventional logical symbols as "A ⇒ ¬B", "B ⇒ ¬A", "¬A ∨ ¬B" and "¬(AB)" respectively.

You can either prove the equivalence of these statements using various deduction rules or just verify that they all have the same truth table:

A ¬A B ¬B A ⇒ ¬B B ⇒ ¬A ¬A ∨ ¬B ¬(AB)
true false true false false false false false
true false false true true true true true
false true true false true true true true
false true false true true true true true

Thus, putting all of this together, "if N is reduced, neither R nor S is reduced" entails "if N is reduced, R is not reduced", which is equivalent to "N and R are not both reduced".

Ps. Perhaps your confusion arises simply from misparsing the statement "N and R are not both reduced" (i.e. "¬(AB)" using the definitions of A and B above) as "N and R are both not reduced" (i.e. "¬A ∧ ¬B" using the same definitions above). These two statements are indeed not equivalent, with the latter being a much stronger claim.

If so, however, this isn't really an issue of logic, but merely one of English grammar, which does have some ambiguity regarding sentences featuring both negation using "not" and conjunction using "both".

In my opinion, the specific word order in the original solution (and in my statement 4 above) isn't actually ambiguous, since in the phrase "not both" it is clear that the negation by "not" must apply to the conjunction introduced by "both", and not the other way around.

However, similar statements using only slightly different word order, such as "both N and R are not reduced", are indeed ambiguous and could be understood either as "¬(AB)" or as "¬A ∧ ¬B" depending on context (and possibly the reader's prior expectations). This makes it very easy to misread such sentences if one isn't careful.

And I suspect that some people might even take issue with my claim that "not both" is unambiguous, and claim that in their dialect of English it can mean the same as "both not".

All that said, in the context of the given solution, it should be clear how the combination of "not" and "both" is meant to be understood, since only one interpretation leads to a logically consistent argument. But of course, to see this, you need to at least briefly consider both interpretations before rejecting the obviously absurd one.

In fact (as pointed out by Ben Voigt in the comments below), the example solution itself seems to fall victim to this grammatical confusion when it concludes that (emphasis mine):

"L and N are a pair of areas that cannot both be reduced if both M and R are reduced"

I at least would formally parse this statement, as written, as "(M is reducedR is reduced) ⇒ ¬(L is reducedN is reduced)", which is a far weaker statement than what the solution actually proves (and needs to prove!).

The proper conclusion, of course, should be "(M is reducedR is reduced) ⇒ (L is not reducedN is not reduced)", which could've been expressed in English e.g. as:

"L and N are a pair of areas that each cannot be reduced if both M and R are reduced"

or perhaps (more closely paralleling the way the question is phrased):

"L and N are a pair of areas neither of which can be reduced if both M and R are reduced"

or even simply:

"neither of L and N can be reduced if both M and R are reduced."

The language is probably more clearly stated as "For N to be reduced R and S must not be reduced.", instead of "If N is reduced, neither R nor S is reduced.", which I read as, "If it is true that N is reduced then it is true that R is not reduced and S is not reduced."

In fact, "For N to be reduced R and S must not be reduced" and "If it is true that N is reduced then it is true that R is not reduced and S is not reduced" are equivalent statements. Indeed, written using logical connectives, both have the exact same form: "N is reduced ⇒ (R is not reducedS is not reduced)".

The only difference is whether, in English, you read the logical implication "AB" as "for A to be true, B must be true" rather than "if A is true, then B is true". But these are just two synonymous ways to state the same logical assertion in English.

Perhaps you're trying to make some distinction about causality with the different phrasings? Or perhaps, given you other remarks such as:

I think this question just doesn't hinge on propositional logic or modes ponens in the way I expected it to.

you might simply have an unreasonably narrow understanding of how propositional logic works.

Indeed, if you limit your available inference rules to pure classical modus ponens (and few if any axioms not given directly in the problem statement), it's quite possible that those tools will be insufficient to solve the problem. For example, while modus ponens allows you to deduce B from A and AB, you will need modus tollens to go the other way from ¬B and AB to ¬A. Or, alternatively, you would need some way to convert AB to its contrapositive ¬B ⇒ ¬A, to which you can then apply modus ponens.

Technically one can prove all valid results of propositional logic using only modus ponens, if one just adds enough logical axioms to emulate all the other inference rules. But that does not seem to be what you're trying to do, either.

Instead, it seems as if you're just trying to apply pure modus ponens directly to the propositions given in the problem statement (not all of which are even in the correct form for it!) without any additional logical axioms (or with just an unstated and apparently quite limited set of such axioms). This is a very weak system of logic (possibly even weaker than minimal logic, depending on what axioms you allow) and unable to prove anything more than the most trivial of syllogisms.

• "misparsing the statement "N and R are not both reduced" (i.e. "¬(A ∧ B)" using the definitions of A and B above) as "N and R are both not reduced" (i.e. "¬A ∧ ¬B" using the same definitions above). These two statements are indeed not equivalent, with the latter being a much stronger claim." is absolutely correct, and the wording of the question requires the stronger statement (the answer wrongly proves the weaker statement, due to unnecessarily weakening its conclusions -- the same argument does lead to the stronger statement also) Commented Jul 26 at 15:20
• @BenVoigt: The stronger statement (i.e. "N and R are both not reduced") can not be deduced from the premises given in the exercise, and indeed contradicts them (since one of the premises given in "question 1" is that "R is reduced"). The weaker statement ("N and R are not both reduced") is sufficient to deduce the desired result ("N is not reduced") from the premises. Commented Jul 26 at 15:29
• Oh right, the statement needed is "M and R are reduced => L is not reduced AND N is not reduced". The "correct answer" in the exercise does wrongly end with a conclusion of "¬(L ∧ N)" when the exercise actually requires that "¬L ∧ ¬N". But I confused that with your analysis of N and R. Sorry for the noise. (On the other hand, you might want to address that rather than telling OP there's no issue). Commented Jul 26 at 17:00
• @BenVoigt: Ah, I see what you mean now. The use of "cannot both be reduced" in the example solution indeed seems like a mistake to me. I added a note about that to my answer. Commented Jul 26 at 18:40

Edit:

To clarify my own confusion, so that if others see this and follow the same path I did they may be able to course correct easier, I was missing a few things.

The fallacy I was expecting here was affirming the consequent, so take this series of statements

``````A = "It is morning"
B = "There are birds chirping"
C = "I am eating dinner"
D = "It is not morning"
``````

And for whatever reason where I live every morning there are birds chirping so `A -> B`, or "If it is morning, there are birds chirping." Suppose I also add that I only eat dinner after 2pm, which is not morning. So, `C -> D`, or, "If I am eating dinner, it is not morning".

Now let's say I am given that there are not birds chirping, ~B. Since a conditional statement is equivalent it's a contrapositive, and A -> B, then ~B -> ~A. So it is not morning. This is the logic I could've used to deduce that, as others have pointed out, `N -> ~R ^ ~S` also means `R or S -> not N`

The fallacy would've been if, say, I tried to reason something like, since I know `C -> D`, or, "If I am eating dinner it is not morning", that it being "not morning" means "I am eating dinner", or `~D -> C`. The contrapositive of `C -> D` is valid, `~D -> ~C`, which is, "If it is morning, then I am not eating dinner".

So, using the contrapositive, I can get some information about the truth of C, not complete information about the situation, but enough to inform an additional inference. I guess this is probably a bit foreign to me being out of academia for so long and following my instincts to look for more intuitive / complete, or general information about what's going on.

I guess the lesson is I probably should brush up on my formal logic if I want to take the LSAT lol.

The below is incorrect and based in misunderstanding.

I think this question just doesn't hinge on propositional logic or modes ponens in the way I expected it to. I also think the question is necessarily ambiguous given my interpretation of it seems valid from a purely mathematical / logical perspective, although I also can see the natural / legal interpretation of the language.

The response to my question describing it as a "condition" seems to capture the essence of it most clearly in my view.

The language is probably more clearly stated as "For N to be reduced R and S must not be reduced.", instead of "If N is reduced, neither R nor S is reduced.", which I read as, "If it is true that N is reduced then it is true that R is not reduced and S is not reduced."

It isn't trying to capture the inference that

A = "N is reduced" B = "Neither R nor S is reduced"

A -> B

Rather it's capturing a condition of the situation such that for N to be reduced R and S must not as a rule.

• The alternate phrasing, which you discard, is equivalent to the story problem. Your proposed phrasing is ambiguous (may mean the same thing, or something weaker) Commented Jul 26 at 15:24
• I'm trying to understand the difference between the inference (implication) that you say it isn't trying to capture, and the condition you say that it is capturing. They sound equivalent to me. Maybe you're making a distinction about the direction of causality? But implication in propositional logic doesn't say anything about causality. A -> B is completely equivalent to ~B -> ~A; neither form says that the truth value of A or B (necessarily) causes the truth value of the other. They both just say which sets of truth values can coexist and which can't. Commented Jul 26 at 16:34

causative provided the essential and formal logic. However, as a slight extension, i would recommend for people to simply walk the graph. I find contra-positive and the like do become more apparent, and not enough people use graphs!

if solid lines indicate affective reductions, and dotted lines indicate immunities. we can read that N must be safe because R has been reduced.

"... if N is reduced, neither R nor S is reduced. So N and R cannot both be reduced."

I don't think an answer has directly mentioned why this statement, though not well-explained, is indeed correct.

First, use the perhaps-unintuitive fact that A => B is logically equivalent to "(not A) or B". (To convince yourself, write down the truth table.) So here, "N reduced => R not reduced" is logically equivalent to "N not reduced or R not reduced".

Next, we note that "not A or not B" is logically equivalent to "not (A and B)". So here, "N not reduced or R not reduced" is logically equivalent to "N and R cannot both be reduced".