Question

The two premises and the conclusion are below, together with my interpretations:

P1) ∀s∃t(¬R(s,t) ∧ ¬(s=t)) - For every element s in the domain, there is a t which R doesn't relate to s and that is not s itself. This means all elements in the domain are not related to themselves and there is one element apart from themselves that they are not related to.

P2) ∀s∃tR(s,t) - for all elements s, there exists a t such that R relates s to t. The conclusion is C) ⊢ ∃s¬R(s,s) - There exists an element that R does not relate to itself. Based on my interpretation above, I think the following demonstrates this: Domain:0,1,2 R(,):<0,1>,<1,0>,<2,0>

I think that based on my interpretation, I think this is correct but I keep getting a response that I am wrong? Or am I meant to provide a counter example?

Answer 1

The student's example meets the conditions set by the premises P1 and P2 and demonstrates the conclusion C that there exists an element not related to itself.

Step-by-step explanation:

The student has interpreted two premises involving a relationship R and is trying to establish the truth of a given conclusion. The first premise P1 states that for every element s, there is another element t such that s is not related to t by R, and t is not the same as s. The second premise P2 states that for every element s, there exists an element t such that s is related to t by R. The conclusion C claims there is at least one element that is not related to itself. The student's example domain {0,1,2} with relations <0,1>, <1,0>, <2,0> indeed demonstrates the conclusion, as it shows an instance where no element is related to itself, and hence, there exists an element (in fact, all in this case) that are not related to themselves, fulfilling the conclusion C.