Proof of Irreflexivity of the Strict Order and Subtracting Zero in ℝ

Proof of (i). By The Real Numbers (i), $\mathbb{R}$ is a set, and by The Structure of ℝ as a Field and an Ordered Ring (iii), $\le$ is a partial order on $\mathbb{R}$; $<$ is its associated strict relation. So Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (iii), applied with $a=\mathbb{R}$ and $u=x\in\mathbb{R}$, gives that $x<x$ does not hold.

Proof of (ii). By Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (i), $0\in\mathbb{R}$, so $-0\in\mathbb{R}$, and hence $x+(-0)\in\mathbb{R}$. By Negatives, Differences, Reciprocals and Quotients (i), $x-0=x+(-0)$, so $x-0\in\mathbb{R}$. By The Structure of ℝ as a Field and an Ordered Ring (viii), applied with $y=0$, $(x-0)+0=x$. By Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v), applied to the real number $x-0$, $(x-0)+0=x-0$. Combining the two equalities, $x-0=x$.