# Proof

Let $$x,y,z$$ be arbitrary natural numbers.

### $$(i)$$ "$$\Rightarrow$$"

According to the definition of order relation of natural numbers, we there exists a natural number $$u\neq 0$$ with $$y=x+u$$. By virtue of the associativity and commutativity laws for adding natural numbers Therefore we have $y+z=(x+u)+z=x+(u+z)=x+(z+u)=(x+z)+u.$

It follows, $$x+z < y + z$$.

### $$(ii)$$ "$$\Leftarrow$$"

Assume, $$x + z < y + z$$ , but not $$x < y$$. According to the trichotomy of the order relation for natural numbers, we must have otherwise $$x = y$$ or $$x > y$$. If $$x = y$$, it would follow from the cancellation law for adding natural numbers that $$x + z = y + z$$, which is a contradiction to the assumption $$x + z < y + z$$. If $$x > y$$, then it would exist a natural number $$v\neq 0$$ with $$x=y+v$$. Then we would get $$(y+v) + z < y + z$$, or equivalently (applying the associativity and commutativity of adding natural numbers) $$(y+z) +v < y+z$$. This is again a contradiction, since $$(y + z) + v > y + z$$. Thus, we must have $$x < y$$.

#### $$(1a)$$ Proof of $$x < y\Longleftrightarrow z + x < z + y$$

Follows from $$(1)$$ and the commutativity of adding natural numbers.

#### $$(2)$$ Proof of $$x > y\Longleftrightarrow x + z > y + z$$ and of $$x > y\Longleftrightarrow z + x > z + x$$

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

#### $$(3)$$ Proof of $$x \le y\Longleftrightarrow x + z \le y + z$$ and of $$x \le y\Longleftrightarrow z + x \le z + y$$

In the "$$<$$" case, the proof is identical to the proof $$(1)$$ or $$(1a)$$, for symmetry reasons. For the "$$=$$" case, the proof is identical to the proof of the cancellation law for adding natural numbers.

#### $$(4)$$ Proof of $$x \ge y\Longleftrightarrow x + z \ge y + z$$ and of $$x \ge y\Longleftrightarrow z + x \ge z + y$$

The proof is analogous to the proof of $$(3)$$, for symmetry reasons.

Github: ### References

#### Bibliography

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