I tried to do the following exercise from a book I'm reading, can you tell me if my solution is correct? Thanks!
Theorem 2: Let $R$ be a binary relation on a set $X$. Then $R$ is strictly well-founded iff there is no sequence $(x_n)_{n \in \mathbb N}$ of elements of $X$ such that $(x_{n+1}, x_n) \in R$ for all $n \in \mathbb N$.
Exercise:
(i) Formulate an analogue of Theorem 2 for well-founded relations.
(ii) Prove Theorem 2.
(i) Theorem 2': Let $R$ be a binary relation on a set $X$. Then $R$ is well-founded iff there is no sequence $(x_n)_{n \in \mathbb N}$ of elements of $X$ such that $(x_{n+1}, x_n) \in R$ and $x_n \neq x_{n+1}$ for all $n \in \mathbb N$.
(ii)
$\implies$: Assume there is a sequence $x_n$ with $(x_{n+1},x_n) \in R$ for all $n \in \mathbb N$. Then the set $Y = \{x_n \mid n \in \mathbb N\}$ and hence $R$ is not well-founded since for every element $x_n$ there is $x_{n+1} \in Y$ with $(x_{n+1},x_n) \in R$.
$\Longleftarrow$: Let $R$ be strictly well-founded. Assume we have a sequence $x_n$ with $(x_{n+1},x_n) \in R$ for all $n \in \mathbb N$. Then by the same argument as above $R$ is not well-founded, hence there cannot be such a sequence.
I must be missing something but I can't seem to figure out what.