Some people are Montagues and some people are Capulets.No Montague can be crossed in love.All Capulets can be crossed...

Naz October 17, 2014

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.

JenaeHB October 22, 2014

Helps clears a lotz

JenaeHB October 22, 2014

Helps clear a lot of questions

Advaith August 10, 2015

Very helpful

RyanSpencer July 15, 2019

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

Ravi July 31, 2019

@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!