# The relation “is strictly higher than” is considered antisymmetric?

I'm studying from Michael Carter's "Foundations" and in the answer key to exercise 1.15 he says that with regard to mountain peaks the relation "is strictly higher than" is antisymmetric. In other words,

if $xRy$ and $yRx$ implies $x=y$

How can the relation be considered antisymmetric if $xRy$ and $yRx$ can't both hold at the same time? Is it a mistake?

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The statement " x doesn't exist => x has the property P" is considered true for all P... In other words if something doesn't exist we can say anything about it.... – N. S. May 13 '11 at 4:57
Thanks! I appreciate the help. – jefflovejapan May 13 '11 at 5:10

What this means is that the condition of anti-symmetry is an implication. The only way for an implication to be false is if the antecedent holds but the consequent does not. Here, in order to show that the relation is not anti-symmetric, you would need to exhibit two elements $x$ and $y$ for which both $xRy$ and $yRx$ are true, but the consequent, $x=y$, is false.
However, as you note, it is impossible to find two elements for which $xRy$ and $yRx$ both hold; that means that the implication can never be false; which means the implication is always true. Which means, of course, that the relation is anti-symmetric.
"By vacuity" means "because it is empty". It comes from the fact that the set of all situations in which the antecedent is true is empty; and when the antecedent cannot be true, the implication is always true. It's the same reason why we can truthfully say "Every element of the empty set is green"; and "Every element of the empty set is blue"; both statements are true by vacuity, because they are of the form "if $x$ is in the empty set, then blah".
In short: the statement "if $P$ then $Q$" is considered to be true if $P$ is always false. This is what we have here: $P$ ("$xRy$ and $yRx$") is always false, so the implication is always true.