Determining non-isomorphic models of a formula 
Let $Q$ be the formula arising from the conjunction of the following formulas (of first-order classical logic with equality):
  
  
*
  
*$\exists x P(x)$
  
*$\exists x \neg P(x)$
  
*$\forall xy (P(x) \wedge P(y) \rightarrow x = y)$
  
*$\forall xy (\neg P(x) \wedge \neg P(y) \rightarrow x = y)$
  
  
  How many models does $Q$ have? How many of them are non isomorphic?

I'm having trouble with counting or giving a formal proof regarding the models of $Q$. Also, I don't exactly get what it means with non-isomorphic models. 
 A: Let's first see what the axioms say. 
(i) The first axiom says there is an object that has property $P$. More formally, in any model $M$ of the axiom system, there is an object $a$ such that $P_M(a)$ is true in $M$. 
(ii) The second axiom says there is an object which doesn't have property $P$. 
(iii) The third axiom says there is no more than one object with property $P$. It does so by saying that if the objects $x$ and $y$ have property $P$, then $x$ must be the same as $y$.
(iv) The fourth axiom says there  is no more than one object that doesn't have property $P$. 
So any model of the full set of $4$ axioms has exactly two elements, one of which has property $P$, and the other of which doesn't. And any two element structure $M$, where $P$ is interpreted to be true at one of the elements and false at the other, is a model.
There are infinitely many models. The underlying set of the model could be $\{1,2\}$, or $\{17,24\}$, or $\{a,b\}$ where $a$ is the set of integers and $b$ the set of reals. Any two-element set can be made the underlying set of a model, by interpreting $P$ to be true at one of the objects and false at the other.
However, up to isomorphism there is only one model. More precisely, if $M$ and $M'$ are any two models, then $M$ and $M'$ are isomorphic.
For let $M=\{a,b\}$, and $M'=\{a',b'\}$, where $P_M(a)$ holds and $P_{M'}(a')$ holds. Then the mapping $\varphi$ that takes $a$ to $a'$ and $b$ to $b'$ is a (structure-preserving) isomorphism from $M$ to $M'$.
