# Prove that a binary relation R over a set A is a strict order if and only if R is irreflexive and transitive

I was wondering if someone can check this proof for me. And if you have any other methods of proving this please feel free to write them below.

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Lemma 1: If $$R$$ over a set $$A$$ is a strict order then $$R$$ is irreflexive and transitive

Proof : Lets assume $$R$$ is a arbitary binary relation over the set $$A$$, is strict order. We need to prove $$R$$ is irreflexive and transitive.

Consequently we already know $$R$$ is a strict order relation. Therefore $$R$$ is irreflexive and transitive.

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Lemma 2: If $$R$$ over a set $$A$$ is irreflexive and transitive then $$R$$ is strict order.

Proof: Lets assume $$R$$ is an arbitrary relation that is reflexive and transitive. We will prove that $$R$$ is strict order.

Consequently, we know $$R$$ is irreflexive and transitive. Therefore we will need to prove $$R$$ is asymmetric.

Let $$a$$ and $$b$$ be arbitrary elements where $$aRb$$ holds. We will prove that $$bRa$$ doesn't hold. Consider $$c$$ to be an element such that $$c=a$$. The relation $$R$$ is transitive we know that $$aRb \cap bRc$$ should result in a false to prove the asymmetry of $$R$$. We know that $$R$$ is irreflexive therefore $$aRc$$ is false.

Notice that $$aRb \cap bRc \implies aRc$$ is true therefore we can conclude that aRb is asymmetric.

• Looks good to me, except that you don't need to introduce $c$ in your second lemma. – Don Thousand Jan 22 '20 at 21:35

• Antisymmetry and asymmetry have different meanings. Antisymmetry is $x\prec y$ and $y\prec x\to x=y$ and asymmetry is $x\prec y\to$ not $y\prec x$. Technically, they are the same in irreflexive relations, but one normally associates antisymmetry with lax orders and asymmetry with strict orders. – Matthew Daly Jan 22 '20 at 23:17