TITLE: Prove that formula is satisfiable in some infinite structure, but invalid in all finite structures
QUESTION [7 upvotes]: Given formula.
$$[\forall x \exists y P(x,y) ]\;\land [\forall x \forall y (P(x,y) \to \lnot P(y,x))] \;\land [\forall x \forall y \forall z ((P(x,y) \land (P(y,z)) \to P(x,z)))]$$
First of all, I tried to find some (infinite) interpretation, where this formula becomes true. While looking at this formula, I noticed that the second part of it is anti-symmetric relation, and the third part is transitivity. So I immediately thought about choosing P as a partial order on natural numbers.
Let $M = \mathbb{N}$ and $P(x,y) := x < y$.
If I choose $y = x + 1$ in the first part, then the formula evaluates to T
OK, if everything is good till this moment, then let's try to prove that the statement is identically F for any finite interpretation.
I'm stuck for a few hours in the second part. Had an idea to blindly 'rename' each sub-formula by some letter, and try to prove it using propositional calculus methods (proof by contradiction or just via truth tables). But the first one $\forall x \exists y P(x,y)$ which involves $\exists$ quantifier confused me.
What did I missed here? Doing self study of material is very time consumable when you're stuck. An answer with detailed explanation will be much valuable.
Thank you.

REPLY [2 votes]: If $\Bbb M$ is any interpretation let $P^M=\{(x,y)\in M: \Bbb M\text { satisfies } P(x,y)\}$ then $P^M$ is a binary relation on $M.\;$(I.e. $P^M$ is a subset/subclass of $M\times M.$). 
Let $\Bbb M$ be a finite interpretation with $n$ members ($n\in \Bbb N$). Suppose $P^M$ satifies the first condition (I.e. the domain of $P^M$ is $M$) and that $P^M$ is transitive. Then $P^M$ cannot be anti-symmetric because $\exists x\;(P^M(x,x)).$ 
Proof: Take $x_1\in M$ and for $1\leq j\leq n$ let $x_{j+1}\in \{y\in M: P^M(x_j,y)\}.$ There exist $i,j$ with $1\leq i<j\leq n+1$ with $x_i=x_j$ by the Pigeon-Hole Principle.  Now  transitivity of $P^M$ implies, by induction on $k$, that  $P^M(x_i,x_{i+k})$ whenever $1\leq k\leq n+1-i.$  But then we have $P^M(x_i,x_j)$ and $x_i=x_j,$ contradicting anti-symmetry.
Remark: If you prefer, let $M=\{m_i:1\leq i\leq n\},$ and let $x_1=m_1$ and let $x_{j+1}=m_u(j)$ where $u(j)=\min \{v: P^M(x_j,m_v\}$, if you want a  recursive algorithm for defining  each $x_j$.