first order logic sentence I'm trying to show that the sentence
$$\forall_x \exists_y P(x,y) \land \forall_x \forall_y (P(x,y) \implies \neg P(y,x)) \land \forall_x \forall_y \forall_z (P(x,y) \implies (P(y,z)\implies P(x,z))$$ (where $P$ is binary predicate) is false in finite structures and true in some infinite structure.
For the second task I took structure $<\mathbb{Z}, < >$ and it seems to work. But i have no idea how to do the first part of the exercise.
 A: What it says is that we have an "strict order" (last part says transitive, second says antisymmetric) on the set where every element has a larger element (which is strictly larger as $P(x,x)$ can never hold, due to part 2 of the statement).
So if $F$ is a finite (non-empty) model, take $f_0 \in F$, and by recursion we find $f_n$ with $P(f_n, f_{n+1})$ for all $n \ge 0$. By finiteness some $f_n = f_m$ for $n \neq m$. Find a contradiction with transitivity.
A: Your sentence is basically made up of three statements.
The proof should go something like:


*

*Start with some $x_0$ in your finite structure

*Because of the first statement, there exists some $x_1$ such that $P(x_0, x_1)$. And some $x_2$ such that $P(x_1,x_2)$. This way, you can define $x_n$ for each $n\in\mathbb N$.

*Obviously, since the structure is finite, there exists some repetition. Take the first $k$ for which $x_k$ appears twice in the sequence $(x_0,x_1,\dots)$, i.e. $x_k=x_{k+m}$ for some $m$

*Now, you have a chain $x_k, x_{k+1},\dots x_{k+m}$.

*Because of the second statement, you know that $P(x_{k+m-1}, x_{k+m})$.

*Because of the third statement, you know that (proove it by induction) $P(x_k, x_{k+m-1})$.

