Equality of Ordered Tuples

Theorem
Let $a = \left({a_1, a_2, \ldots, a_n}\right)$ and $b = \left({b_1, b_2, \ldots, b_n}\right)$ be ordered tuples.

Then:
 * $a = b \iff \forall i: 1 \le i \le n: a_i = b_i$

That is, for two ordered tuples to be equal, all the corresponding elements have to be equal.

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\left({a_1, a_2, \ldots, a_n}\right) = \left({b_1, b_2, \ldots, b_n}\right) \iff \forall i: 1 \le i \le n: a_i = b_i$

$P(1)$ is true, as this just says $\left({a_1}\right) = \left({b_1}\right) \iff a_1 = b_1$ which is trivial.

Basis for the Induction
$P(2)$ is the case:
 * $\left({a_1, a_2}\right) = \left({b_1, b_2}\right) \iff a_1 = b_1, a_2 = b_2$

which has been proved in Equality of Ordered Pairs.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:
 * $\left({a_1, a_2, \ldots, a_k}\right) = \left({b_1, b_2, \ldots, b_k}\right) \iff \forall i: 1 \le i \le n: a_i = b_i$

Then we need to show:
 * $\left({a_1, a_2, \ldots, a_{k + 1}}\right) = \left({b_1, b_2, \ldots, b_{k + 1}}\right) \iff \forall i: 1 \le i \le n: a_i = b_i$

Induction Step
This is our induction step:

But from the induction hypothesis we have that:
 * $\left({a_2, \ldots, a_{k + 1}}\right) = \left({b_2, \ldots, b_{k + 1}}\right) \iff \forall i: 1 \le 2 \le k + 1: a_i = b_i$

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N_{>0}: \left({a_1, a_2, \ldots, a_n}\right) = \left({b_1, b_2, \ldots, b_n}\right) \iff \forall i: 1 \le i \le n: a_i = b_i$