Proving transitivity of the negation of a partial order relation

Question

Let S be a finite set with the partial order $\preceq$, where $ a\preceq b \iff (a=b \lor k(a) < k(b)) $ for $k(x) \in K, K$ is a set with a total order $\leq$.

What I want to prove is that $\forall a,b,c \in S$:

$(a\npreceq b \land b \npreceq c \land a \neq c)\implies a \npreceq c, $

By negation I get $x \npreceq y \iff (x \neq y \land k(x) \nless k(y))$, thus I have

$a \npreceq b \iff (a \neq b \land k(a) \nless k(b))$

$b \npreceq c \iff (b \neq c \land k(b) \nless k(c))$

To show that

$a \npreceq c \iff (a \neq c \land k(a) \nless k(c))$

I already have $a \neq c$, thus I need to show that $(k(a) \nless k(b) \land k(b) \nless k(c))\implies k(a) \nless k(c)$. How do I conclude that?

Answer 1

Recall that the order on $K$ is total. Thus $\not<$ is actually equivalent to $\geq$ for this order. Then the thing you are trying to prove is precisely transitivity of $\geq$.