Prove that $\neg \left [ \exists ! x \in \mathbb{R} \exists ! c \in \mathbb{R} (x^2 + 3x + c = 0) \right ]$ This is an exercise from Velleman's "How To Prove It". I have been struggling with the logical form of this statement, and just want to make sure that my proof is correct now.

Show that it is not the case that there is a unique real number $x$ such that there is a unique real number $c$ such that $x^2 + 3x + c = 0$. (Hint: You should be able to prove that for every real number $x$ there is a unique real number $c$ such that $x^2 + 3x + c= 0$

$$\neg \left [ \exists ! x \in \mathbb{R} \exists ! c \in \mathbb{R} (x^2 + 3x + c = 0)  \right ]$$
$$\neg \left [ \exists x \in \mathbb{R} \left(\exists ! c \in \mathbb{R} (x^2 + 3x + c = 0) \wedge \neg \exists y \in \mathbb{R}(\exists ! c \in \mathbb{R} (x^2 + 3x + c = 0) \wedge x \neq y)\right )  \right ]$$
$$ \forall x \in \mathbb{R} \left( \neg \exists ! c \in \mathbb{R} (x^2 + 3x + c = 0) \vee \exists y \in \mathbb{R}(\exists ! c \in \mathbb{R} (x^2 + 3x + c = 0) \wedge x \neq y)\right ) $$
$$ \forall x \in \mathbb{R} \left( \exists ! c \in \mathbb{R} (x^2 + 3x + c = 0) \rightarrow \exists y \in \mathbb{R}(\exists ! c \in \mathbb{R} (x^2 + 3x + c = 0) \wedge x \neq y)\right ) $$
Proof: Let $x \in \mathbb{R}$ be arbitrary. Suppose that $\exists ! c \in \mathbb{R} (x^2 + 3x + c = 0) $. Let y = x + 1. Clearly $x \neq y$. Now let $c =  - x^2 -5x - 4$.
Then
\begin{align}
y^2 + 3y + c &=(x+1)^2 + 3(x+1) - x^2 - 5x -4 \\ 
&= x^2 + 2x + 1 + 3x + 3 - x^2 -5x - 4 \\
&= 0
\end{align}
To show that $z$ is unique, let $d \in \mathbb{R}$ be arbitrary such that $y^2 + 3y + d = 0$. Then subtracting $y^2 + 3y$ from both sides of the equation yields $d = -y^2 - 3y = -(x+1)^2 - 3(x+1) = -x^2 -5x - 4 = c. \square$
 A: Indeed do this by contradiction is just a one line proof, but what you did is still correct as a direct proof, if we write as following might make it easier to read.
Let $P(x)\equiv\exists ! c \in \mathbb{R}, (x^2 + 3x + c = 0) $
\begin{align}
&\neg(\exists ! x \in \mathbb{R},P(x))\\
\equiv&\exists^{>1}x\in\mathbb{R},P(x)\lor \forall x\in\mathbb{R},\neg P(x)\\
\equiv&\exists x\in\mathbb{R},P(x)\to\exists^{>1}x\in\mathbb{R},P(x)
\end{align}
Assume $\exists x\in\mathbb{R},P(x)$, have $\exists x\in\mathbb{R},\exists ! c \in \mathbb{R}, x^2 + 3x + c = 0$, fix $x$ and since that $$(x+1)^2+3(x+1)=x^2+5x+4$$
take $c_0=c-2x-4$, this proves $(x+1)^2+3(x+1)+c_0=0$, where $c_0$ is unique.
A: Following the hint, one can show that in fact, the statement holds after removing $\neg$ and the square brackets, and replacing the first $\exists!$ with $\forall$.
For any real number $x$, there is obviously a unique real number $c$ for which $x^2+3x+c=0$, namely $c=-x^2-3x$.
Hence, the statement "$\forall x \in \mathbb{R}\, \exists! c \in \mathbb{R}\, x^2+3x+c=0$" is true, and so certainly, "$\exists! x \in \mathbb{R}\, \exists! c \in \mathbb{R}\, x^2+3x+c=0$" is false.
