The Wikipedia article on WQOs does not mention a *strict* version. I came across a particular relation, which I could only describe as a strict WQO, but I am wondering if my reasoning is correct and if the notion of a “strict WQO” even makes sense (I think it does).

Consider the set $ X \times \mathbb{N}$ where $ X = \{1,\ldots,k\}$ for some integer $ k$ and $ \mathbb{N}$ denotes the set of natural numbers. We define the relation $ \prec$ as: $ $ (x_1,n_1) \prec (x_2,n_2) \quad \text{iff} \quad (x_1 \leq x_2) \wedge (n_1 < n_2) $ $

*Strict Quasi Order*: Clearly $ \prec$ is irreflexive and transitive, and therefore a strict quasi order.

*No Infinite Antichains*: It’s also easy to observe that there are no infinite antichains. Due to the Pigeonhole principle, any sequence $ (x_1,n_1), (x_2,n_2), \ldots$ of length $ \geq k$ must have at least one pair $ \langle (x_i,n_i),(x_j,n_j)\rangle$ such that $ x_i = x_j$ . Then, since either $ n_i < n_j$ or $ n_j < n_i$ must hold, we have at least one pair in the sequence on which $ \prec$ holds.

*Well-Foundedness*: Finally, we show that there are no infinite descending chains. For any element $ (i,j)$ , there are only $ |\{1,\ldots,i\} \times \{1,\ldots,(j-1)\}| = i(j-1)$ elements that can precede $ (i,j)$ .

Is this reasoning correct? Then is $ \prec$ a strict WQO on $ X \times \mathbb{N}$ ?

**More questions:** Since antisymmetry is implied by irreflexivity and transitivity, is any strict quasi order also a strict partial order? And so is any strict WQO also a strict WPO?

Thanks.