1. ## Finding a counterexample

Find a domain for the quantifiers in $\displaystyle \exists x\exists y(x\neq y \wedge \forall z(z\neq x \wedge z\neq y))$ such that this statement is true.

2. I doubt it is possible to find the same domain for all three quantifiers. How can one avoid instantiating z, in particular, to x?

In fact, I used my favorite proof assistant program to prove the negation of this statement. The actual code (the proof is omitted) is the following.

Code:
Parameter T : Type.

Theorem t :
~ (exists x : T, exists y : T, (x <> y /\ forall z : T, (z <> x /\ z <> y))).
Proof. ... Qed.

3. Yeah, I was able to prove it after a couple of tries as well, on this website: The Logic Machine: Logic Software at Texas A&M University
|-~$x$y(~x=y&@z(~z=x&~z=y))
OK 1 (1) $x$y(~x=y&@z(~z=x&~z=y)) A
OK 2 (2) $y(~a=y&@z(~z=a&~z=y)) A OK 3 (3) ~a=b&@z(~z=a&~z=b) A OK 3 (4) @z(~z=a&~z=b) 3&E OK 3 (5) ~a=a&~a=b 4@E OK 3 (6) ~a=a 5&E OK (7) a=a =I OK 3 (8) ~$y(~a=y&@z(~z=a&~z=y)) 6,7RAA(2)
OK 2 (9) ~$y(~a=y&@z(~z=a&~z=y)) 2,8$E(3)
OK 2 (10) ~$x$y(~x=y&@z(~z=x&~z=y)) 2,9RAA(1)
OK 1 (11) ~$x$y(~x=y&@z(~z=x&~z=y)) 1,10$E(2) OK (12) ~$x\$y(~x=y&@z(~z=x&~z=y)) 1,11RAA(1)