Finite model to demonstrate invalidity of semantic consequence

Question

I am attempting to show that $\exists xRxx$ is not a semantic consequence of $\forall x \forall y \forall z[(Rxy\land Ryz)\Rightarrow Rxz]$ and $\forall x \exists yRxy$. In coming up with a model to show this, my instinct is the natural numbers with the < relation; everything is less than something, "less than" is transitive, and yet nothing is less than itself. Is there a model with a finite number of terms that could demonstrate this?

Answer 1

Your example with the natural numbers indeed shows that $\exists x Rxx$ is not a semantic consequence of the two other statements. However, given a model $M$ of those two statements and $\not\exists x Rxx$, it should be straightforward to use the relation $R$ to cook up a surjection from $M$ onto the natural numbers, which shows that $M$ is not finite.

Answer 2

There is no such finite model. Suppose $M$ is one, with universe $\{m_1, ... m_n\}$, and suppose $x\in M$. By induction, define $x_0 = x$, and $x_{n+1} =$ some $y\in M$ such that $xRy$. By transitivity of $R$, $x_i R x_j$ if $i<j$. Because $M$ is finite, there must be $i\ne j$ such that $x_i = x_j$. Then $x = x_i$ is such that $xRx$.