In a Mathematical Introduction to Logic, Enderton says it is easy to see (by using induction on $n$ and the basic property of ordered pairs ($\left\langle x,y\right\rangle =\left\langle u,v\right\rangle \iff x=u\land y=v$ )) that (for all $n$) if $\left\langle x_{1},\ldots,x_{n}\right\rangle =\left\langle y_{1},\ldots,y_{n}\right\rangle$ then $x_{i}=y_{i}$ for $1\leq i\leq n$. I'm having some difficulty doing this
Definitions
Enderton defines $\left\langle x\right\rangle:= x$ so the base case $n=1$ is trivial. He also defines $\left\langle x_{1},\ldots,x_{n+1}\right\rangle :=\left\langle \left\langle x_{1},\ldots,x_{n}\right\rangle ,x_{n+1}\right\rangle$.
Attempt:
For the induction step, we want to show $\left\langle x_{1},\ldots,x_{n}\right\rangle =\left\langle y_{1},\ldots,y_{n}\right\rangle \rightarrow \left\langle x_{1},\ldots,x_{n+1}\right\rangle =\left\langle y_{1},\ldots,y_{n+1}\right\rangle $
By definition, we have
$\begin{align} \left\langle x_{1},\ldots,x_{n+1}\right\rangle &=\left\langle y_{1},\ldots,y_{n+1}\right\rangle \\ \left\langle \left\langle x_{1},\ldots,x_{n}\right\rangle ,x_{n+1}\right\rangle &=\left\langle \left\langle y_{1},\ldots,y_{n}\right\rangle ,y_{n+1}\right\rangle \end{align}$
but the basic property of ordered pairs is that
$\left\langle x,y\right\rangle =\left\langle u,v\right\rangle \iff x=u\land y=v$
Thus $x_{n+1}=y_{n+1}$ and $\left\langle x_{1},\ldots,x_{n}\right\rangle =\left\langle y_{1},\ldots,y_{n}\right\rangle$
So I've proved $$ \left\langle x_{1},\ldots,x_{n+1}\right\rangle =\left\langle y_{1},\ldots,y_{n+1}\right\rangle \iff x_{n+1}=y_{n+1} \land \left\langle x_{1},\ldots,x_{n}\right\rangle =\left\langle y_{1},\ldots,y_{n}\right\rangle$$
the induction hypothesis is $\left\langle x_{1},\ldots,x_{n}\right\rangle =\left\langle y_{1},\ldots,y_{n}\right\rangle$, but I don't see how to derive the other premise: $x_{n+1}=y_{n+1}$, or make the induction step.
Attempt #2 For the induction hypothesis, we assume there exists $n\in\mathbb{N}$ such that:
$\left\langle x_{1},\ldots,x_{n}\right\rangle =\left\langle y_{1},\ldots,y_{n}\right\rangle \rightarrow x_{i}=y_{i}(1\leq i\leq n)$
In the induction step, we show that this implies:
$\left\langle x_{1},\ldots,x_{n+1}\right\rangle =\left\langle y_{1},\ldots,y_{n+1}\right\rangle \rightarrow x_{i}=y_{i}(1\leq i\leq n+1)$.
Thus for the induction step, we may assume
$\begin{align} \left\langle x_{1},\ldots,x_{n+1}\right\rangle &=\left\langle y_{1},\ldots,y_{n+1}\right\rangle \\ \left\langle \left\langle x_{1},\ldots,x_{n}\right\rangle ,x_{n+1}\right\rangle &=\left\langle \left\langle y_{1},\ldots,y_{n}\right\rangle ,y_{n+1}\right\rangle \end{align}$
but the defining property of ordered pairs is that
$\left\langle x,y\right\rangle =\left\langle u,v\right\rangle \iff x=u\land y=v$.
Thus $x_{n+1}=y_{n+1}$ and $\left\langle x_{1},\ldots,x_{n}\right\rangle =\left\langle y_{1},\ldots,y_{n}\right\rangle$ . Thus by the induction hypothesis, we have $x_{i}=y_{i}(1\leq i\leq n)$. Since we showed $x_{n+1}=y_{n+1}$, we have $x_{i}=y_{i}(1\leq i\leq n+1)$ which is what we wanted to prove.
Your goal for the "induction step" is the wrong way around. Since $\left<x_1,\ldots\right>=\left<y_1,\ldots\right>$ is a premise of the implication you want to prove, the right induction step should go as:
Our goal is to show that if $\left<x_1,\ldots,x_{n+1}\right> =\left<y_1,\ldots,y_{n+1}\right>$ then all of the $x_i$s and $y_i$ agree.
Thus, assume $\left<x_1,\ldots,x_{n+1}\right> =\left<y_1,\ldots,y_{n+1}\right>$.
The induction hypothesis tells us that if $\left<x_1,\ldots,x_n\right> =\left<y_1,\ldots,y_n\right>$ then the first $n$ values agree.
Now, if we can show $$\left<x_1,\ldots,x_{n+1}\right> =\left<y_1,\ldots,y_{n+1}\right> \to \left<x_1,\ldots,x_n\right> =\left<y_1,\ldots,y_n\right>,$$ then together with (2) that will give us what whe need to apply the induction hypothesis. ...