# Show that a set is consistent (first-order logic)

Check if $A$ is consistent or not:

$A = \{ \forall x \exists y \exists z(Pxy \land Pzx), \\ \forall x, \forall y, \forall z(Pxy \rightarrow Pyz \rightarrow Pxz) \\ \forall x \forall y(Pxy \rightarrow \neg Pyx) \\ \forall x \forall y (x\neg\approx y) \rightarrow (Pxy \lor Pyx)) \}$

I am not exactly sure how to go about doing this problem. I started by setting $D$ as my model with domain $\{a, b, c \}$, but I don't know if this is the right approach, or how to proceed.

• Should the last two "$z$"s be "$x$"s? Apr 28 '17 at 0:26
• Yes, I corrected that now. Apr 28 '17 at 0:36
• I suspect that you've incorrectly transcribed the second and third sentences. Check that you've correctly copied those down. Apr 28 '17 at 0:37
• It should be correct now Apr 28 '17 at 0:38
• These properties have names: unboundedness, transitivity, anti-reflexivity, and totality. Hope that helps. Apr 28 '17 at 0:55

The second, third, and fourth statement express that $P$ is a 2-place relation that is transitive, asymmetric, and total. Together, that makes $P$ a total order, meaning that all objects in the domain can be put from 'left' to 'right', or from 'smaller' to 'bigger'. Moreover, the first statement says that every object has an object to the right of it/bigger than it, as well as to the left of it/smaller than it, implying that there is no 'leftmost'/'smallest' or 'rightmost'/'biggest' object. And that means that a finite domain will never satisfy these sentences.
However, it is now easy to come up with an infinite-sized model for these sentences: just take as the domain all the integers, and let $Pxy$ mean $x<y$.