Proof of Adding a Positive Real, Cancelling an Added Real, and Regrouping Four Summands in ℝ

Throughout, all sums and negatives formed below are real numbers, by Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (i). Since $\mathbb{R}$ is an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (ii), Rules of Arithmetic and Order in an Ordered Field applies to it.

(i): Suppose $0<y$. Adding $x$ to both sides preserves the strict inequality by Rules of Arithmetic and Order in an Ordered Field (iv), so $0+x<y+x$. Here $0+x=x$ by Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v) and $y+x=x+y$ by Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (ii), hence $x<x+y$.

(ii): By the definition of the difference, then Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (iii), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (vi) and Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v) in turn,

$$(x+y)-y=(x+y)+(-y)=x+\bigl(y+(-y)\bigr)=x+0=x.$$

(iii): Using Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (iii) in the first, second, fourth and fifth steps and Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (ii) (for $b+c=c+b$) in the third,

$$(a+b)+(c+d)=a+\bigl(b+(c+d)\bigr)=a+\bigl((b+c)+d\bigr)=a+\bigl((c+b)+d\bigr)=a+\bigl(c+(b+d)\bigr)=(a+c)+(b+d).$$