# Proof

### $$(i)$$ Proof of the transitivity law $$x < y\wedge y < z\Longrightarrow x < z$$.

Assume that for any natural numbers $$x,y,z\in\mathbb N$$, the order relations * $$x < y$$ and $$y < z$$ are fulfilled. Then there exist natural numbers $$u\neq 0$$ and $$v\neq 0$$ with $$y=x+u$$ and $$z=y+v$$. From the associativity of adding natural numbers, it follows

$z=y+v=(x+u)+v=x+(u+v).$ Since $$u+v\neq 0$$, it follows $$x < z$$.

### $$(ii)$$ Proof of the transitivity law $$x \le y\wedge y \le z\Longrightarrow x \le z$$.

It follows from $$(i)$$ and the transitivity of the equality relation "$$=$$".

### $$(iii)$$ Proof of the transitivity law $$x > y\wedge y > z\Longrightarrow x > z$$.

Analogously to $$(i)$$, there exist natural numbers $$u\neq 0$$ and $$v\neq 0$$ with $$x=y+u$$ and $$y=z+v$$. From the associativity of adding natural numbers, it follows

$x=y+u=(z+v)+u=z+(v+u).$ Since $$v+u\neq 0$$, it follows $$x > z$$.

### $$(iv)$$ Proof of the transitivity law $$x \ge y\wedge y \ge z\Longrightarrow x \ge z$$.

It follows from $$(iii)$$ and the transitivity of the equality relation "$$=$$".

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### References

#### Bibliography

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