Proof

$$(i)$$ Proof of "$$\Rightarrow$$"

If $$y > x$$, then it follows from the definition of order relation that there exist a natural number $$u\neq 0$$ with $$y=x+u$$. According to the fact that every natural number is greater or equal zero, we must have $$u > 0$$. Because $$1=0^+$$ is the unique successor of $$0$$, it is also the smallest number $$u$$, for which $$u > 0$$. Therefore, if $$u=1$$, then $$y = x +1$$. Otherwise $$y > x+1$$. Altogether, it follows $$y \ge x+1$$.

$$(ii)$$ Proof of "$$\Leftarrow$$"

Assume $$y \ge x +1$$ and $$y > x +1$$. Then we have $$x + 1 > x$$ and by the transitivity of the order relation, $$y > x$$.

Assume $$y \ge x +1$$ and $$y = x +1$$. Then we have $$x + 1 > x$$ and we have trivially $$y > x$$.

$$(2)$$ Proof of $$y < x\Longleftrightarrow y + 1 \le x$$

The proof is analogous to the proof of $$(1)$$, for symmetry reasons.

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References

Bibliography

1. Landau, Edmund: "Grundlagen der Analysis", Heldermann Verlag, 2008