Order of Real Numbers is Dual of Order of their Negatives/Proof 2

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Theorem

$\forall x, y \in \R: x > y \iff \paren {-x}< \paren {-y}$


Proof

\(\ds x\) \(<\) \(\ds y\)
\(\ds \leadstoandfrom \ \ \) \(\ds y - x\) \(>\) \(\ds 0\) Inequality iff Difference is Positive
\(\ds \leadstoandfrom \ \ \) \(\ds \) \(>\) \(\ds 0\) Real Number Axioms: $\R \text A 2 $: Commutativity of Addition
\(\ds \leadstoandfrom \ \ \) \(\ds -x + -\paren {-y}\) \(>\) \(\ds 0\) Negative of Negative Real Number
\(\ds \leadstoandfrom \ \ \) \(\ds -x - \paren {-y}\) \(>\) \(\ds 0\) Definition of Real Subtraction
\(\ds \leadstoandfrom \ \ \) \(\ds -y\) \(<\) \(\ds -x\) Inequality iff Difference is Positive

$\blacksquare$