$\forall x,y,z,u\in\mathbb{N}, (x\geq y \wedge z>u)\vee (x>y\wedge z\geq u)\Rightarrow xz>yu$ is true for all natural numbers? Even zero?

Question

$\forall x,y,z,u\in\mathbb{N}, (x\geq y \wedge z>u)\vee (x>y\wedge z\geq u)\Rightarrow xz>yu$ is true for all natural numbers? Even zero?

I have come to the conclusion that this property is fulfilled by all natural numbers, but I still do not know how to argue that zero also fulfills this property. Could someone please tell me why zero also fulfills this property? Thank you very much!

Answer 1

Let $x=y=0$ and $z=1, u=0$, then $xz > yu$ is false.

Your definition of $\mathbb{N}$ excluded $0$ if this statement is claimed to be true.

Answer 2

Consider the case x=1, y=0, u=0, z=0:

$x\geq y \wedge z>u$ is false but

$x>y\wedge z\geq u$ is true, so the antecedent is true.

But $xz \gt yu$ is false.

So the implication is false when you allow $u=0$