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 tuples to be equal, all the corresponding elements have to be equal.

Proof
Proof by induction:

For all $$n \in \N^*$$, 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:
 * $$\forall n \in \N^*: \left({a_1, a_2, \ldots, a_n}\right) = \left({b_1, b_2, \ldots, b_n}\right) \iff \forall i: 1 \le i \le n: a_i = b_i$$