Proposition 2.2.12: Lat $a$,$b$,$c$ be natural numbers. Then
(a) $a ≥ a$,
(b) If $a≥b$ and $b≥c$, then $a≥c$,
(c)If $a≥b$ and $b≥a$, then $a=b$,
(d) $a≥b$ if and only if $a+c≥b+c$,
(e) $a<b$ if and only if $a++≤b$, here $a++$ shows successor of a,
(f)$a<b$ iff $b=a+d$ for some positive natural number d.
Note: This is not complete proof, the latter half i have understood almost. But have confusion in former part.
My understanding : Tao splitted the proof in two parts. In first part he shall show at a time we have only one statement out of three. And in second part he shall show atleast one statement can be true. Thus overall he showed exactly one statement will be true for any natural numbers a and b. So in first part if $a<b$ is true, then $a≠b$. Similarly if $a>b$, then $a≠b$. This shows all three statements can't exist at a same time. Now he has to show no two statements out of 3 also can't exist at a same time.
Questions:
(1) How Tao showed no two statements holds at a time?
(2) "If $a<b$ and $b<a$, then by proposition 2.2.12 we have a=b, a contradiction". I really didn't understand this line. How " $a<b$ and $b<a$ $\implies$ $a=b$ ?" How we got contradiction?
(3) Is my above understanding correctly? Please need feedback.
(4) Is there any other way to prove this proposition. Without using Tao's method given in the book?
Thank You.