It also seemed to me not to be a wff, but I wasn't sure. Kmath, I meant that I couldn't myself translate the logical statement in your first post into a grammatically correct one.

I looked over the second example in the text that I alluded to earlier, probably for the 20th time, and then just took it all for granted. The following is that example, and then after that is my new solution patterned after this example, which I will have to reflect on a little more so that I feel that I understand it sufficiently.

**The Example**

2. Analyze the logical form of the statement $\displaystyle \{x_i\ |\ i \in I\} \subseteq A$ by writing out the definitions of the set theory notation used.

*Solution*. By the definition of subset we must say that every element of $\displaystyle \{x_i\ |\ i \in I\}$ is also an element of $\displaystyle A$, so we could start by writing $\displaystyle \forall x(x \in \{x_i\ |\ i \in I\} \rightarrow x \in A)$. Filling in the meaning of $\displaystyle x \in \{x_i\ |\ i \in I\}$, which we worked out earlier, we would end up with $\displaystyle \forall x(\exists i \in I(x = x_i) \rightarrow x \in A)$. But since the elements of $\displaystyle \{x_i\ |\ i \in I\}$ are just the $\displaystyle x_i$'s, for all $\displaystyle i \in I$, perhaps an easier way of saying that every element of $\displaystyle \{x_i\ |\ i \in I\}$ is an element of $\displaystyle A$ would be $\displaystyle \forall i \in I(x_i \in A)$. The two answers we have given are equivalent, but showing this would require the methods we will be studying in Chapter 3.

**My New Solution**

If Example 2 says, in regard to a certain set being a subset of another set, that "every element of $\displaystyle \{x_i\ |\ i \in I\}$ is an element of $\displaystyle A$ would be [to say] $\displaystyle \forall i \in I(x_i \in A)$," then the logical form of the statement $\displaystyle \{n^2 + n + 1\ |\ n \in \mathbb{N}\} \subseteq \{2n + 1\ |\ n \in \mathbb{N}\}$ can be analyzed as

$\displaystyle \equiv \forall n \in \mathbb{N} (n^2 + n + 1 \in \{2m + 1\ |\ m \in \mathbb{N}\})$

$\displaystyle \equiv \forall n \in \mathbb{N} (n^2 + n + 1 \in \{x\ |\ \exists m \in \mathbb{N} (x = 2m + 1)\})$

$\displaystyle \equiv \forall n \in \mathbb{N} (\exists m \in \mathbb{N} (n^2 + n + 1 = 2m + 1))$

$\displaystyle \equiv \forall n \in \mathbb{N} \exists m \in \mathbb{N} (n^2 + n + 1 = 2m + 1)$.