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Given a binary relation R,S on set A, assume that R is anti-symmetric. Show R intersection S is anti-symmetric.

I started this proof by stating the definition of anti-symmetric with R which is $$ ∀a∀b(((a, b) ∈ R ∧ (b, a) ∈ R) → (a = b))$$ I know the basic idea that since R does not have any elements that make it not anti-symmetric, (i.e. for any $(a,b) ∈ R$ (b,a) is not an element of R), any set $R ∩ S$ will not include elements that would make R not anti-symmetric because elements of the intersections have to include both sets. Because those same elements that would make R not antisymmetric any are not included in R, they would not be in $R ∩ S$. Even if S was not antisymmetric and it had elements (a,b) and (b, a) where a does not equal b, and (a,b) is common with R, $R ∩ S$ would not have have (b,a) thus making it antisymmetric. I know these things but I do not know how to write them into a formal proof. I am stuck at this.

user278039
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2 Answers2

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It's best if you write down the property that $R\cap S$ must satisfy. You want to prove that $R\cap S$ is antisymmetric, so you want to show that

$$\forall a,b:((a,b)\in R\cap S \land (b,a)\in R\cap S\implies (a=b))$$

Now, you prove this claim like any other claim starting with $\forall$:

Let $a,b$ be such that $(a,b)\in R\cap S \land (b,a)\in R\cap S$. Something something something Therefore, $a=b$.

5xum
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I think the confusing thing is trying to look at relation $S$ separately. Suppose we use a new letter $T$ for the intersection $R \cap S.$ Then if $(a,b)$ is in $T$ and also $(b,a)$ is in $T,$ it follows that both $(a,b)$ and $(b,a)$ are in $R$ because $T \subseteq R,$ because $T=R \cap S.$ From this, because $R$ is antisymmetric, we arrive at $a=b.$ Thus $T$ is antisymmetric as desired.

coffeemath
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