Proposition:
Suppose $R$ is a partial order on a set $A$. Prove that every finite, nonempty set $B ⊆ A$ has an $R$-minimal element.
My attempt:
Notation:
If $A$ is a set and $A = \{3,5\}$, then its cardinality will be denoted as
$$|A| = 2 $$
By induction.
Base case:
Suppose $B \subseteq A$ and $|B| = 1$. Then clearly, the only element $B$ has will also be its minimal element.
Induction step:
Take some $B' \subseteq A$, such that $|B'| = n$, and $B'$ has a minimal element, say $c$.
Take $z \in A$ such that $z \notin B'$ and let $B = B' \cup \{z\}$.
If $cRz$, then minimal element of $B$ will be $c$
Suppose $zRc$. Suppose exists $a \in B'$ such that $aRz$. Then by transitivity we have $aRc$. Since $c$ is minimal element, $a = c$. Then it follows that $cRz$ and $z = c$, but this is impossible. Hence provided that $zRc$, $z$ will be the minimal element of $B$.
If $\lnot cRz$ and $\lnot zRc$, then $c$ will be the minimal element of $B$.
$\Box$
Is it correct?