Do we have to show that the relation is irreflexive?

[evinda]

Hello! (Smile)

• Let $R$ be an order of the set $A$. Then $R$ induces a strict order $S$ at the set $A$.
  $$$$
• Let $S$ be a strict order of the set $A$. Then $S$ induces an order $R$ at the set $A$.

The first sentence is proven like that in my notes:

We define $S:=R-I_A$ and we can see that $S$ is a strict order at $A$.

$$x \in A\\y \in A$$

$$xSy \leftrightarrow \langle x,y \rangle \in S=R-I_A\\ \langle x,y \rangle \in R \wedge \langle x,y \rangle \notin I_A$$

We want to show that $\langle y,x \rangle \notin S$.
If not then $\langle y,x \rangle \in S \wedge \langle y,x \rangle \in I_A \leftrightarrow \langle x,x \rangle \notin I_A$.In that way we have shown that $S$ isn't weak antisymmetric.
But don't we also have to show that $S$ is irreflexive?
We know that $S$ is transitive as a subset of a transitive relation, right? (Thinking)

 

[Evgeny.Makarov]

We define $S:=R-I_A$ and we can see want to show that $S$ is a strict order at $A$.

$$x \in A\\y \in A$$

For all $x,y$? For some $x,y$? Which ones? Please use words.

$$xSy \leftrightarrow \langle x,y \rangle \in S=R-I_A\\ \langle x,y \rangle \in R \wedge \langle x,y \rangle \notin I_A$$

OK, I am being picky, but what is the relationship of the second line to the first?

We want to show that $\langle y,x \rangle \notin S$.

Why do we want to show that?

If not then $\langle y,x \rangle \in S \wedge \langle y,x \rangle \in I_A \leftrightarrow \langle x,x \rangle \notin I_A$.

There are too many ways to parse this.

(1) $\langle y,x \rangle \in S$, and also $\langle y,x \rangle \in I_A$ is equivalent to $\langle x,x \rangle \notin I_A$ (but both sides of the equivalence may be false).

(2) $\langle y,x \rangle \in S$ and $\langle y,x \rangle \in I_A$ are together equivalent to $\langle x,x \rangle \notin I_A$ (but both sides may be false).

(3) We have $\langle y,x \rangle \in S$ and $\langle y,x \rangle \in I_A$, and since these statements together are equivalent to $\langle x,x \rangle \notin I_A$, we have the latter.

There are possibly other ways. Please stop abbreviating the flow of reasoning with arrows. This is not helping. Please write words.

In that way we have shown that $S$ isn't weak antisymmetric.

I am not familiar with this concept.

But don't we also have to show that $S$ is irreflexive?

Yes.

We know that $S$ is transitive as a subset of a transitive relation, right?

Not every subset of a transitive relation is transitive.