Suggestion:

Accept $ R $ is an overall order on the set $ A $, Prove that every finite,

Non-empty sentence $ B subseteq A $ has a $ R $-minimal element.

My attempt:

Cardinality of the crowd, we say $ A $is described as $ | A | $,

By induction.

**Base case:**

To take $ B = {b } $ so that $ b in A $, $ b $ is the $ R $-minimal element of $ B $,

**Induction step:**

Suppose, for all sentences with $ n $ Elements, it will be the smallest element.

Consider any quantity $ B subseteq A $ so that $ | B | = n + 1 $,

Any take $ b in B $,

To let $ B = B setminus {b } $, After the inductive hypothesis $ B & # 39; $ Has $ R $-minimal element. Let's call it $ c $,

Let's look at set $ B $,

Accept $ cRb $, Then $ c $ is the smallest element of $ B $,

Accept $ bRc $, Any take $ y in B $, We know that $ cRy $, Through transitivity we have $ bRy $, $ y $ was therefore arbitrary $ b $ is the smallest element of $ B $,

Suppose that $ lnot bRc $ and $ lnot cRb $, That's impossible because $ R $ is a total order on $ A $,

Therefore not $ c $ or $ b $ becomes the smallest element of $ B $,

$ Box $

Is it right?