Definition of antisymmetric binary-relation is

$$\forall a,b\in\mathrm{A},\left[ \left(aRb\wedge bRa\right)\rightarrow\left(a=b\right)\right].$$

Let $\mathrm{A}=\left\{a\mid a\in\mathbb{R}\right\}$, and the relation set $R=\left\{\left(a,b\right)\mid a,b\in\mathbb{R}\wedge a\text{ is odd}\wedge b\text{ is even}\right\}.$

And $R$ is a antisymmetric binary-relation on $\mathrm{A}$. However, I don't understand.

For example, $2R2$ is false and $2=2$ is true. Then the logical expression becomes $\mathrm{F}\wedge\mathrm{F}\rightarrow\mathrm{T}$.

What am I missing? Is my concept on logical expression wrong?

  • 2
    $\begingroup$ Hint: $A \to B$ is true iff $B$ is true or $A$ is false. $\endgroup$ – Jean-Claude Arbaut Nov 29 '13 at 0:05
  • $\begingroup$ I do understand another equivalent definition, $$\forall a,b\in\mathrm{A},\left[ \left(aRb\wedge a\neq b\right)\rightarrow\left(\neg bRa\right)\right].$$ $\endgroup$ – Novice Nov 29 '13 at 0:19

Your evaluation that the antisymmetry condition has, in this case, a false antecedent and a true consequent is correct. But remember that a conditional will only be false if those truth values are the other way around; if you have $\mathbf{T}\to\mathbf{F}$. In this case, for antisymmetry to fail, we'd need an $x,y$ with $xRy\wedge yRx\wedge x\neq y$. But then both $x$ and $y$ would need to be both even and odd, which none are (much less two are), so no contradiction of antisymmetry is possible here.


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