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Proof

(related to Corollary: Rules of Calculations with Inequalities)

Proof of Rule 1) For any $x\neq 0$ it is $x \cdot x >0$. This is a consequence from the rules of multiplying positive and negative real numbers. Proof of Rule 2) $1 > 0$. From the uniqueness of 1 and the uniqueness of 0, it follows that $1 \neq 0$. Therefore, according to the definition of order relation, it must be either $1 > 0$ or $-1 > 0$. Since $1\cdot 1 =1$ , it must be $1 > 0$, according to the rule Rule 1).

Proof of Rule 3)  From $x > y$ it follows that $x+a > y+a$ for any $a\in\mathbb R$. (The rule is analogously valid and proven for any inequalities with "$<$" in between). $x > y$ means, by definition, that $x - y > 0$. Therefore, it is $x - y + a - a > 0$, and so it follows that $x + a  - (y + a) > 0$, and finally that $x + a > y + a$.  

Proof of Rule 4)  If $x > y$ and $a > b$, then it follows that $x+a > y+b$. (The rule is analogously valid and proven for any inequalities with "$<$" in between). By definition, it is $x - y > 0$ and $a - b > 0$. From the definition of order relation, it follows that $(x - y) + (a -b) > 0$, which by definition results in $x + a > y + b$.

Proof of Rule 5) Transitivity of the "$<$" (analogously the "$>$" relation) If $x < y$ and $y < z$, then by definition it is $0 < y - x$ and $0 < z - y$. From the definition of order relation, it follows that $0 < y - x + z - y= z - x$, which by definition results in $x < z$.

Proof of Rule 6) The inequality $x > y$ does not change, if it is multiplied by any number $a > 0$, i.e. it is $ax > ay$. (The rule is analogously valid and proven for an inequalities with "$<$" in between). By definition, it is $x -y  > 0$. From the definition of order relation, it follows that $a(x-y)  = ax - ay > 0$, which, by definition, results in $ax > ay$.

Proof of Rule 7) The inequality $x > y$ changes, if it is multiplied by any number $a < 0$, i.e. it is $ax < ay$. (The rule is analogously valid and proven for an inequality with "$<$" in between, which then changes into "$>$"). By definition, it is $x - y  > 0$ and $-a > 0$. From the definition of order relation, it follows that $(-a)(x-y)  = -ax + ay > 0$, which, by definition, results in $ay > ax$, and finally in $ax < ay$.

Proof of Rule 8)  If $x > 0$, then $x^{-1} > 0$. It is $x^{-1}=x^{-1}\cdot 1=x^{-1}\cdot(x^{-1}\cdot x) =(x^{-1}\cdot x^{-1})\cdot x$, which is greater then $0$, following the precondition, the Rule 1) and the definition of order relation.

Proof of Rule 9)  If $x < 0$, then $x^{-1} < 0$. It is $x^{-1}=x^{-1}\cdot 1=x^{-1}\cdot(x^{-1}\cdot x) =(x^{-1}\cdot x^{-1})\cdot x$, which is smaller then $0$, following the precondition, the Rule 1) and the Rule 7).

Proof of Rule 10) If $y > x > 0$, then $x^{-1} > y^{-1}$. (The rule is analogously valid and proven for the inequality $0 < x < y$, resulting in $y^{-1} < x^{-1}$). According to the precondition $x$ and $y$ are both positive. Therefore, it follows from the definition of order relation that $xy > 0$, and according to Rule 8) $(xy)^{-1} > 0$. Multiplying the inequality $y > x$ by the number $(xy)^{-1}$ does not change the inequality by Rule 6), therefore $y(xy)^{-1} > x (xy)^{-1}$. This is equivalent to $x^{-1} > y^{-1}$.

Proof of Rule 11) If $0 \le x < y$ and $0 \le a < b$  then $ax < by$. (The rule is analogously valid and proven for $by > ax$, following from $y > x \ge 0$ and $b > a \ge 0$). The argument is valid, if $x=0$ or $a=0$, since then $0 < by$ follows from the definition of order relation. So assume $0 < x < y$ and $0 < a < b$. In this case we have $ax < ay$ and $ay < by$, following in both cases from the Rule 6. The argument follows now from the transitivity (Rule 5)).

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