Twisted

Let's diagram: "Some people are Montagues," P1: P-some-M M-some-P "and some people are Capulets." P2: P-some-C C-some-P "No Montague can be crossed in love." P3: M ==> not CL CL ==> not M "All Capulets can be crossed in love." P4: C ==> CL not CL ==> not C "Anyone who is not a Montague is intemperate." P5: not M ==> I not I ==> M "no Montague is intemperate" P6: M ==> not I I ==> not M "Therefore, Capulets are not Montagues." C: C ==> not M M ==> not C Answer choice (E): "All Capulets are intemperate." (E): C ==> I not I ==> not C We can connect P4 to the contrapositive of P3 like so: C ==> CL ==> not M to conclude: C ==> not M, which we can connect to P5 like so: C ==> not M ==> I. Therefore, we can conclude: C ==> I, which is answer choice (E). Hope that clears things up! Please let us know if you have any other questions.

Helps clears a lotz

Helps clear a lot of questions

Very helpful

So, this questions focuses on our ability to use the transitive property?

@JenaehB and @Advaith, we're happy to read that this was helpful! @RyanSpencer, this question tests one's ability to understand quantifiers and put together conditional logic chains that can push out inferences. Let us know if you have any other questions!