Where did I go wrong? Analyze the logical form of $\{n^2+n+1 | n \in \mathbb{N}\} \subseteq \{2n+1 | n \in \mathbb{N}\}$ This question is taken from Velleman's $\textit{How to Prove it}$. It is in the exercises section of 2.3, Question 1c:
$\{n^2+n+1 | n \in \mathbb{N}\} \subseteq \{2n+1 | n \in \mathbb{N}\}$
My work is as follows,
The statement is equivalent to $\forall x(x \in \{n^2+n+1 | n \in \mathbb{N}\} \to x \in \{2n+1 | n \in \mathbb{N}\})$ (definition of a subset)
$x \in \{n^2+n+1 | n \in \mathbb{N}\} \equiv \exists n \in \mathbb{N}(x=n^2+n+1)$
and
$x \in \{2n+1 | n \in \mathbb{N}\} \equiv \exists n \in \mathbb{N}(x = 2n+1)$
so the final expression ends up as,
$\forall x (\exists n \in \mathbb{N}(x=n^2+n+1) \to \exists n \in \mathbb{N}(x=2n+1))$
The solution provided is $\forall n \in \mathbb{N} \: \exists m \in \mathbb{N}(n^2+n+1=2m+1)$ which makes perfect sense to me looking at it in retrospect.
I guess the overarching concern that I have is the methodology involved in solving the question - my approach was to break the statement down into smaller parts and then to rewrite after interpreting each individual segment (not an unreasonable strategy IMO, that seems to be what Velleman has been advocating up to this point). Was that approach incorrectly applied in this situation? Or did I make a technical error that I am just not aware of?
Thanks in advance for any help/insight that can be given!

There is no error. Your solution is correct too. Two things, however:


*

*Although what you wrote is correct, it is always a good idea not to use the same symbol ($n$, in your case) for two different purposes. So, I would have written$$\forall x (\exists n \in \mathbb{N}(x=n^2+n+1) \to \exists m \in \mathbb{N}(x=2m+1))$$

*The provided solution is also correct, but shorter and easier to understand because there is no need to use the symbol $x$. It's as if you had written “for every number, if it belongs to the first set, then it also belongs to the second one” and the proposed solution was “every element of the first set belongs to the second one”.

