At Most Two Distinct Members of A The quantified predicate logic statement that describes at most two distinct members of A, where A, is some arbitrary set is:
$\forall$xyz( (Px $\land$ Py $\land$ Pz) $\Rightarrow$ (x=y $\lor$ x=z $\lor$ y=z) )
I can parse this quantified statement into three cases:


*

*There are No Members of A

*There is 1 Member of A

*There are 2 Members of A
In case 2, x must equal y, then in turn for any z candidate, z is either equal to x, or to y, but x=y.
(So all three "or" statements are true.)
In case 3, x does not equal y, therefore for any z, z will either be equal to x, or z will be equal to y. (The first "or" fails, but the the other two are true.)
In case 1 I don't see how the quantified predicate statement guarantees that no membership can exist, i.e. there are non such members of A. If we don't choose any x, y, z then there are no such x,y, or z in A and the quantified statement "short circuits"? How do I interpret this or reason about it?
 A: In the case 1 (no elements) the antecedent $(Px ∧ Py ∧ Pz)$ is  is false and the implication is true.
A: P is a property that defines A.
For Case 1:
Objects x,y, and z exist in the Universe but since (Px∧Py∧Pz) was false, no candidate was held true for the property P that defines A thus their existence has no identity in A because they are not members of A; therefore, A is empty if the antecedent is false always -the consequent has no effect on defining membership in A. 
The consequent only ensures that IF there is a member in A for no matter how many candidates greater than or equal to one either they either all the same object or they are two distinct objects.
For Case 2:
(Px∧Py∧Pz) is true implying that it must be the case where at least one element is in A. The consequent checks to see what object in A it is- Since this is case 1 x=y and y=z and by transitivity of equivalence x=z. Therefore there is only one distinct member in A for case 1.
For Case 3:
(Px∧Py∧Pz) is true implying that it must be the case that two distinct elements are in A. The consequent checks to see what object(s) in A it is but in this case x $\neq$ y, so x=y check is irrelevant in the disjunctive statement so x and y are distinct and for any other candidate z checked it must be an alias for x or y.  
A: The statement is best read as (using inclusion in a set instead of a predicate; it doesn't really matter):

If $x$, $y$, and $z$ are all in $A$, then there is some pair of them which are equal.

Clearly, this cannot hold for any set $A$ containing three distinct members - hence, we know that the above statement implies that $A$ has size no more than $2$. However, if $A$ has size of $2$ or less, it is equivalent to the pigeonhole principle, when we have $2$ "pigeonholes" and three elements (pigeons?) which we're trying to fit in the holes - and the statement says that it is impossible to assign $x$, $y$, and $z$ to members of $A$ without overlap.
We don't really want to break into cases for this, I don't think. What I just wrote is essentially a proof of "the statement implies $A$ has no more than two elements" followed by a proof of "$A$ having no more than two elements implies the statement" - whereas you were only handling the latter statement, which does not suffice to show the equivalence alone (since it doesn't prove that no set with $3$ elements, for instance, satisfies the condition).
