by ohthatpatrick Wed Mar 06, 2013 7:50 pm
I'm not really sure about what some of your notation means, so let's clarify:
N some O
(Some N's are O's)
-O<---> -P
(Not O if and only if Not P)
-P <---> - Q
(Not P if and only if Not Q)
Q ---> R
(If Q, then R)
We could chain some of those together
~O <---> ~P <---> ~Q
Conversely,
O <---> P <---> Q
We could also use the contrapositive of the last rule to get
~R --> ~Q --> ~P --> ~O
What do we know about N from the first statement?
Some N's = O, P, Q
The first inference you said we had was
N some - P
(Some N's are NOT P)
We can NOT infer that, if I've understood your rules properly. We can infer that Some N's are P, but we can't infer that Some N's are ~P.
R some -P
(Some R's are ~P)
We can NOT infer that, either. We technically know nothing about R, because it's the necessary side of the conditional statement.
If you told me that at least one Q exists, then I know that at least one R exists (since Q-->R). Furthermore, I would know that at least one R is a Q, and if some R's are Q, then some R's are Q, P, and O.
But I still would not be able to infer that some R's are NOT P.
But aren't these also valid inferences
N some - Q (Some N's are ~Q)
I can't infer that. I can only infer that some N's are Q.
N some - R (Some N's are ~R)
I can't infer that. I know that ~R is ~Q and hence ~R is also ~P and ~O. But I can't make inference about ~R and N.
R some - O (Some R's are ~O)
This is the same thing as my explanation for "Some R's are ~P". If you tell me at least one Q exists, then I know some R is Q, and therefore that some R is also P and O. I can't infer that Some R's are NOT O.
Given that none of these inferences were actually valid, I feel like maybe I have misconstrued your notation.
Let me know if you have any clarification about what you were typing or any questions about what I was saying can/can't be inferred.