Your answer is incorrect because we cannot have an S->N statement before the quantifier. Instead of "C->B-some-X," we need "X-some-B->C." The Sufficient condition must be the condition shared by the S->N statement and the quantifier. In other words, the arrow should always point away from the quantifier.
We start with the following: P1: X - most - B P2: ? C: X - most- C
We can then make deductions for the two statements we are given. We know that "most" statements are reversible into "some" statements. This gives us the following: P1: X - most - B B - some - X P2: ? C: X - most - C C - some - X
We also know that some statements are reversible, which gives us the following: P1: X - most - B B - some - X X - some - B P2: ? C: X - most- C C - some - X X - some - C
Now that we have made all of the deductions that we can from our original two statements, we need to consider how to connect P1 to C. How can we get from "X-some-B" to "X-some-C"? These two statements are exactly the same, except B changes to C. We can achieve this with the premise: B -> C. This gives us a chain of: X-some-B->C. The arrow points away from the quantifier, so it is valid. Our contrapositive is: not C -> not B. This gives us a final answer as follows: P1: X - most - B B - some - X X - some - B P2: B -> C not C -> not B C: X - most- C C - some - X X - some - C
Does that make sense? Please let us know if you have any other questions!
gharibiannickMarch 10, 2020
thank you. i believe i got it. i may be revisiting the quantifiers questions and lessons in the future lol
AndreaKMarch 17, 2020
@Gharibiannick, Glad you got it! Feel free to let us know if you have anymore questions about this.
