# Suppose R is a partial order on a set A, and B sube A. Prove that, if R is a total order and b is a minimal element of B, then b is the smallest element of B.

Suppose $R$ is a partial order on a set $A$, and $B\subseteq A$.
Prove that, if $R$ is a total order and $b$ is a minimal element of $B$, then $b$ is the smallest element of $B$.
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Gaige Burton
It's easy. Let $\le$ denote the total order. Let $b$ be any minimal element of $B$, and $u$ another element of $B$ different from $b$. Then $u\le b$ or $b\le u$. If $u\le b$ then you get $u=b$, a contradiction. Thus, $b\le u$, so $b$ is minimum.