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- mcrittell
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Atticus Finch
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Q21 - Not all tenured faculty
This one seems fuzzy; can someone please clear it up for me? Thanks!
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- demetri.blaisdell
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Re: Q21 - Not all tenured faculty
This is a great question. It really tests our understanding of which inferences are valid and which are not. The core is fairly easy to diagram.
Not all tenured faculty are full professors + All linguistics profs have tenure → Some linguistics profs aren’t full professors
The first step on Match the Flaw questions is to identify the flaw. If it doesn’t jump out at you while you diagram the question, it might help to take a closer look at the first premise. "Not all tenured faculty are full professors" is exactly the same thing as saying "at least one tenured professor isn’t a full professor." This statement applies to the university as a whole. When we try to apply this statement to a subgroup (the linguistics department), we have a whole/part problem. All we need to satisfy the premise is at least one professor at the university to be tenured and not a full professor. We can’t infer anything about the linguistics department from that premise. It could be true that the entire linguistics department is composed of full professors.
[By the way, it could also be true that there isn’t a single full professor in the linguistics department. In fact, it’s at least theoretically possible that no tenured faculty member at the entire university is a full professor (every professor is "at least one").] If this bracketed section didn’t make sense, feel free to skip it.
So we’re looking for an answer choice that displays the same whole/part flawed inference. If the conditional logic in the answer choices is tripping you up, diagram them.
(C) gives us the same flaw and it is presented in the same order as the stimulus (thank you, LSAT). Some buildings designed by famous architects are not well proportioned (same meaning as "not all buildings designed by FAs are well proportioned"). All government buildings are designed by famous architects. Therefore, some of them are not well proportioned. The same flaw applies here: maybe famous architects are so patriotic that every government building is well proportioned, but some of their private projects are not well proportioned. The point is that we don’t know which of their buildings are not well proportioned, so we can’t infer anything about a certain category of their buildings.
(A) is not flawed. All modern office buildings have A/C. Some office buildings don’t have A/C. Therefore, some office buildings are not modern (if they were modern, they’d have A/C).
(B) is not flawed. It contains a valid conditional inference. Some municipal hospital buildings are massive and not forbidding. Therefore, not every massive building is forbidding.
(D) contains a very different flaw. Some public buildings are poorly designed. Some of these poorly designed public buildings were intended for private use. Therefore all poorly designed public buildings were intended for private use. This answer choice simply changes the conditional statement from some to all.
(E) also contains a valid inference. It has similar logic to (B). If some cathedrals are not built of stone and are impressive, then non-stone buildings can be impressive.
I hope this explanation helps cut through the fuzziness. Let me know if you have any questions.
Demetri
Not all tenured faculty are full professors + All linguistics profs have tenure → Some linguistics profs aren’t full professors
The first step on Match the Flaw questions is to identify the flaw. If it doesn’t jump out at you while you diagram the question, it might help to take a closer look at the first premise. "Not all tenured faculty are full professors" is exactly the same thing as saying "at least one tenured professor isn’t a full professor." This statement applies to the university as a whole. When we try to apply this statement to a subgroup (the linguistics department), we have a whole/part problem. All we need to satisfy the premise is at least one professor at the university to be tenured and not a full professor. We can’t infer anything about the linguistics department from that premise. It could be true that the entire linguistics department is composed of full professors.
[By the way, it could also be true that there isn’t a single full professor in the linguistics department. In fact, it’s at least theoretically possible that no tenured faculty member at the entire university is a full professor (every professor is "at least one").] If this bracketed section didn’t make sense, feel free to skip it.
So we’re looking for an answer choice that displays the same whole/part flawed inference. If the conditional logic in the answer choices is tripping you up, diagram them.
(C) gives us the same flaw and it is presented in the same order as the stimulus (thank you, LSAT). Some buildings designed by famous architects are not well proportioned (same meaning as "not all buildings designed by FAs are well proportioned"). All government buildings are designed by famous architects. Therefore, some of them are not well proportioned. The same flaw applies here: maybe famous architects are so patriotic that every government building is well proportioned, but some of their private projects are not well proportioned. The point is that we don’t know which of their buildings are not well proportioned, so we can’t infer anything about a certain category of their buildings.
(A) is not flawed. All modern office buildings have A/C. Some office buildings don’t have A/C. Therefore, some office buildings are not modern (if they were modern, they’d have A/C).
(B) is not flawed. It contains a valid conditional inference. Some municipal hospital buildings are massive and not forbidding. Therefore, not every massive building is forbidding.
(D) contains a very different flaw. Some public buildings are poorly designed. Some of these poorly designed public buildings were intended for private use. Therefore all poorly designed public buildings were intended for private use. This answer choice simply changes the conditional statement from some to all.
(E) also contains a valid inference. It has similar logic to (B). If some cathedrals are not built of stone and are impressive, then non-stone buildings can be impressive.
I hope this explanation helps cut through the fuzziness. Let me know if you have any questions.
Demetri
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- WaltGrace1983
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Atticus Finch
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Re: Q21 - Not all tenured faculty
I've been really struggling with parallel flaws with "some" / "most" statements so I thought I'd share my process in learning how to alleviate my struggles. Demetri's explanation rocks but I just thought I'd share a different, more visual, way of getting to the same place. Some people don't like diagramming but I feel like if I spend more front-end time diagramming, I spend less back-end time mulling over wrong answers.
"Not all tenured faculty...full professors."
"Therefore...every faculty member in the linguistics...tenure"
*Notice that you can link these two statements up to form...
"Not all...in linguistics...are full professors."
In a nutshell, here is what is going on:
Makes sense right? No! The statement is not saying SOME L's are ~FP, it is saying SOME TF's are ~FP. That is the critical distinction and one that the original argument is flawed for not making.
(A) some OB → ~CC→ ~MO ⊢ some OB → ~MO
Perfectly logical and I got to it by mapping out the conditional statements and utilizing the contrapositive of (MO → CC).
(B) MH → M, MH → some ~FA ⊢ M → some ~FA
This is flawed but the structure is different from the original argument. The original argument is simply stringing together one long line of conditional logic. This one doesn't involve any stringing together of logic. We could conclude that some MH are ~FA but not anything about M.
(C) GB → FA → some ~WP ⊢ GB → some ~WP
Same exact flaw as the original. Once again, the some ~WP applies ONLY to FA, we don't know if ~WP applies to GB (it might, it might not).
(D) I eliminated this automatically because it doesn't use any "all" statement in the premise, which is necessary for a parallel argument.
(E) C → I, C → some ~BS ⊢ some (I & ~BS)
This is also perfectly valid. We know ALL C's are I and SOME C's are ~BS. Thus, there must be some overlap between I and ~BS. This is what the argument concludes.
"Not all tenured faculty...full professors."
- TF → some ~FP
That is, IF you are tenured faculty (TF), you might be ~FP (not a full professor)
"Therefore...every faculty member in the linguistics...tenure"
- L → TF
Here is where I went wrong the first time. I think you would be best taking this as another premise. It seems to make the most sense that way. Either way, this is a pretty easy conditional statement: IF you are faculty member in lingustiics (L) you will be a tenured faculty member (TF)
*Notice that you can link these two statements up to form...
- L → TF → some ~FP
This says that every faculty member in linguistics in tenured, but some tenured faculty members are NOT full professors. However, we don't know WHO the people are that are NOT full professors. Maybe every tenured faculty in linguistics is a full professor or maybe no tenured faculty in linguistics is a full professor. Maybe some tenured faculty in linguistics are; some aren't.
"Not all...in linguistics...are full professors."
- ⊢ L → some ~FP
In a nutshell, here is what is going on:
- L → TF → some ~FP
⊢ L → some ~FP
Makes sense right? No! The statement is not saying SOME L's are ~FP, it is saying SOME TF's are ~FP. That is the critical distinction and one that the original argument is flawed for not making.
(A) some OB → ~CC→ ~MO ⊢ some OB → ~MO
Perfectly logical and I got to it by mapping out the conditional statements and utilizing the contrapositive of (MO → CC).
(B) MH → M, MH → some ~FA ⊢ M → some ~FA
This is flawed but the structure is different from the original argument. The original argument is simply stringing together one long line of conditional logic. This one doesn't involve any stringing together of logic. We could conclude that some MH are ~FA but not anything about M.
(C) GB → FA → some ~WP ⊢ GB → some ~WP
Same exact flaw as the original. Once again, the some ~WP applies ONLY to FA, we don't know if ~WP applies to GB (it might, it might not).
(D) I eliminated this automatically because it doesn't use any "all" statement in the premise, which is necessary for a parallel argument.
(E) C → I, C → some ~BS ⊢ some (I & ~BS)
This is also perfectly valid. We know ALL C's are I and SOME C's are ~BS. Thus, there must be some overlap between I and ~BS. This is what the argument concludes.
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