Q5 - Elena: While I was at the dog show

[dangilomartin]

So I diagrammed the following:

DG--> WP
WP ---> DG
Combined

DG <---> WP

(A) Never says White dogs but only White poodles, She could have saw white dogs that where not poodles (fails fact test)

(B) Fact test ( not mentioned)

(C) Correct If a dog growled (DG then WP) . It could not have been gray.

(D) Fails fact test, Never stated that White dogs but white poodles ( shell game trick)

(E) Out of Scope, never states this and not a necessary conclusion

Is this the right method of attack. I got this answer wrong under timed conditions when I reviewed it without time this is my mindset. I want to make sure I am getting the right answer for the right reason and eliminating the wrong answer for the right reasons.

[ohthatpatrick]

You nailed it!

This question, like many Inference questions, is testing the two conditional rules supplied. As you notated, we get a bi-conditional statement.

We know that for every dog she saw,
the dog growled and was a white poodle
or
the dog did not growl and was not a white poodle.

I think your reasons were solid (you were using terms I didn't recognize, like Fact Test and Shell Game, but it was easy to understand what you meant by them).

I'll write a quick round of answer choice explanations that describe the wrong answers in terms of "could be otherwise" (the opposite of Must Be True), but your way was fine.

(A) She may have seen tons of other breeds of dogs, but none of those other breeds growled at her.

(B) There may have been gray poodles there ... she might not have seen them ... if she did see them, then they didn't growl at her.

(D) If she saw a white labrador, we know that it did NOT growl at ther.

(E) She may have seen gray poodles, but they didn't growl at her.

Nice work!

[Harmmanb]

What if it was a white poodle that she did not see? The second conditional statement says: every white poodle I saw growled at me.
So you have:

1. Growled -> White poodle

2. White poodle she saw -> Growled.

Or since the sufficient condition is modified/qualified by "being seen" by her, wouldn't you be able to also diagram the second conditional rule as:

White poodle + Seen -> Growled.


For the first statement, she does not need to see the dog to know that it growled at her, but in the second statement, she does for it to be sufficient. Looking at it this way, you wouldn't be able to create a double arrow conditional statement; there could be white poodles that she did not see that did not growl at her. This is impossible if diagrammed with a double arrow.

You still get the same answer with this analysis, I'm wondering if this is correct?
You can get AC C based off of the first condition alone. If it growled at her -> White poodle/Not a white poodle -> Did not growl
Grey dog = Not white -> Did not growl

[ohthatpatrick]

You’re not wrong, but it seems like you’re going too deep into this one.

Anyway, as silly as it is to try defend the biconditional, which is unnecessary, I would say that this statement is dubious: “For the first statement, she does not need to see the dog to know that it growled at her” … sure, but she kinda does, both to verify that the growling is directed at HER, not someone else and to be able to have identified it as a white poodle (unless the growl of a white poodle is a clearly identifiable sound).

So I think you could add in “+ seen” to both rules.

Growled <——> White poodle + seen.