Writing "$f$ is not a bijection" with quantifiers only Is it possible to write '$f$ is NOT a bijection' with quantifiers only, and without using "$\neg$"? What is the negation of "$\exists$!"?
 A: For $f:X \to Y$ we can write something like
$$ \bigl(\exists x_1,x_2\in X:f(x_1)=f(x_2)\land x_1\ne x_2\bigr)
\lor \bigl(\exists y\in Y:\forall x\in X:f(x)\ne y\bigr)$$
But it seems to be hard to avoid the negations implicit in the $\ne$s.
A: Assume that the formal language only has $=$, $f$, variables, and quantifiers. Then there is no formula of the type requested in the question. 
Suppose that $\phi$ is a formula that does not include $\lnot$. I claim that, in a structure with only one element, $\phi$ must be true. But every function from that model to itself is a bijection, so no $\phi$ can express "$f$ is not a bijection" in all models.
To prove that such a formula $\phi$ is true a model with only one element, we just have to consider all the cases, where we only consider formulas that do not include $\lnot$.


*

*Every substitution instance of the formula $x = y$ will be true, because there is only one element. Every atomic formula is a substitution instance of $x=y$.

*Every expression of the form $\psi \land \theta$, $\psi \lor \theta$, or $\psi \Rightarrow \theta$ will be true when $\psi$ and $\theta$ are both true

*Every quantified expression $(\exists x)\psi$ of $(\forall x)\psi$ will be true, because every substitution instance of $\psi$ will be true.
Once we exclude models with only one element, the question has a postive answer. $(\forall x)(\forall y)(x = y)$ is then a suitable replacement for $\bot$. an identically false formula. And then $\lnot \theta$ is $\theta \Rightarrow \bot$ as usual, so we can express negation without using $\lnot$. 
A: Hint: The expression
$$\exists ! x \, P(x)$$
is shorthand for 
$$\exists x \, P(x) \; \land \forall y \, \forall z \, ((P(y) \land P(z)) \implies y=z)$$
which can be negated with deMorgan's laws.
A: $$\exists y\in Y\quad \forall x\in X:\qquad f^{-1}\bigl(\{y\}\bigr)\ne\{x\}\ .$$
