Question: **Prove by induction**.Every **partial order** on a **nonempty finite** set at least one **minimal** element. 

How can **I** solve that question ?

Answer: 
It is trivially true if there is only one element in the poset. Now suppose it is true for all sets of size < n. Compare the nth element to the minimal element of the (n-1) poset, which we know to exist. It will either be the new minimal or not or incomparable. It doesn't matter either way. (Why?)
If the partial order has size 1, it is obvious.

Assume it is true for partial orders `<n`, and then take a partial order `(P,<)` has size `n`.

Pick `x` in `P`. Let `P(<x) = { y in P : y<x }`

If `P(<x)` is empty, then `x` is a minimal element.

Otherwise, `P(<x)` is strictly smaller than `P`, since `x` is not in `P(<x)`. So the poset `(P(<x),<)`
must have a minimal element, `y`.

This `y` must be a minimal element of `P` since, if `z<y` in `P`, then `z<x`, and hence `z` would be in `P(<x)` and smaller than `y`, which contradicts the assumption that `y` was minimal in `P(<x)`.