Find a domain for the quantifiers in such that this statement is true.
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.
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
I don't know why the book asks you to do something that's impossible. Thanks though.|-~$x$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)
Congratulations. Your proof is correct.