a < b if and only if a++ ≤ b. I have to prove that $ a < b$ if and only if $a++\leq  b$.
I am using the book analysis 1 by Terence Tao, which unfortunately has no section for solutions of exercises.
both $ a $ and $b$ are natural numbers, in this case, natural numbers start at $0$.
the notation $a++$ represent the successor of a number, for instance $0++ = 1$
also, $a \leq b$ is defined as  $b = a + d$, for some natural number $d$, and $a < b$ is defined as $a \leq b$ and $a \neq b$
I have tried to use the definition.
If $a < b$, then $a \neq b$ and $b = a + d$, since $a \neq b$, $d > 0$, then I tried to add $1$ to both side and I got $a++ < b++$,
which is obviously not what I have to prove, I have also tried with proof by contradiction but I didn't find any contradiction. Since I couldn't prove it, I tried with the converse theorem, if $a++ \leq b$, then $a < b$, I used the fact that $a < a++$, and $a++ \leq b$, $a \neq a++$, therefore $a \neq b$ and by transitivity, $a < b$, I'm not sure if this is completely correct because I haven't proved yet that $a < a++$, and so far, the book hasn't mentioned that $a < a++$, though it is obvious. the axiom 2(peano's axioms) says that if $a$ is a natural number, then $a++$ is also a natural number. 

Depending on the precise definition of the order, addition, and the properties you know, you can argue as follows:
You have $b=a+d$ for some $d$, with $d\gt 0$. Since $d\gt 0$, it is a successor, so $d=c\mathrm{++}$ for some $c\geq 0$. Thus, 
$$b = a+d = a+(c\mathrm{++}) = (a+c)\mathrm{++} \geq a\mathrm{++}$$
since $a+c\geq a$. 
