Equality of Ordered Tuples

Theorem
Let $a = \tuple {a_1, a_2, \ldots, a_n}$ and $b = \tuple {b_1, b_2, \ldots, b_n}$ 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 $\map P n$ be the proposition:
 * $\tuple {a_1, a_2, \ldots, a_n} = \tuple {b_1, b_2, \ldots, b_n} \iff \forall i: 1 \le i \le n: a_i = b_i$

$\map P 1$ is true, as this just says $\tuple {a_1} = \tuple {b_1} \iff a_1 = b_1$ which is trivial.

Basis for the Induction
$\map P 2$ is the case:
 * $\tuple {a_1, a_2} = \tuple {b_1, b_2} \iff a_1 = b_1, a_2 = b_2$

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 $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

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

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

Induction Step
This is our induction step:

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

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

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