# 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:

 $\ds \tuple {a_1, a_2, \ldots, a_{k + 1} }$ $=$ $\ds \tuple {b_1, b_2, \ldots, b_{k + 1} }$ $\ds \leadstoandfrom \ \$ $\ds \tuple {a_1, \tuple {a_2, \ldots, a_{k + 1} } }$ $=$ $\ds \tuple {b_1, \tuple {b_2, \ldots, b_{k + 1} } }$ $\ds \leadstoandfrom \ \$ $\ds a_1$ $=$ $\ds b_1$ $\ds \land \ \$ $\ds \tuple {a_2, \ldots, a_{k + 1} }$ $=$ $\ds \tuple {b_2, \ldots, b_{k + 1} }$ Basis for the Induction

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:

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

$\blacksquare$

## Examples

### Ordered Triple

Let:

$\tuple {a_1, a_2, a_3}$ and $\tuple {b_1, b_2, b_3}$

Then:

$\tuple {a_1, a_2, a_3} = \tuple {b_1, b_2, b_3}$
$\forall i \in \set {1, 2, 3}: a_i = b_i$