# Minimal WRT Restriction

## Theorem

Let $A$ be a set or class.

Let $\RR$ be a relation on $A$.

Let $B$ be a subset or subclass of $A$.

Let $\RR'$ be the restriction of $\RR$ to $B$.

Let $m \in B$.

Then:

$m$ is a strictly minimal element under $\RR$ in $B$
$m$ is a strictly minimal element under $\RR'$ in $B$.

## Proof

### Sufficient Condition

Let $m$ be a strictly minimal element under $\RR$ in $B$.

Let $x$ be any element of $B$.

Aiming for a contradiction, suppose that $x \mathrel {\RR'} m$.

Then since $\RR' \subseteq \RR$:

$x \mathrel \RR m$

contradicting the fact that $m$ is a strictly minimal element under $\RR$ in $B$.

Thus:

$\lnot \paren {x \mathrel {\RR'} m}$

As this holds for all $x \in B$, $m$ is a strictly minimal element under $\RR'$ in $B$.

$\Box$

### Necessary Condition

Let $m$ be a strictly minimal element under $\RR'$ in $B$.

Let $x \in B$.

Aiming for a contradiction, suppose that $x \mathrel \RR m$.

Then $x, m \in B$.

Therefore:

$\tuple {x, m} \in B \times B$

Thus:

$\tuple {x, m} \in \RR \cap \paren {B \times B} = \RR'$

so $x \mathrel {\RR'} m$

This contradicts the fact that $m$ is a strictly minimal element under $\RR'$ in $B$.

Thus:

$\lnot \paren {x \mathrel \RR m}$

As this holds for all $x \in B$, it follows that $m$ is a strictly minimal element under $\RR$ in $B$.

$\blacksquare$