FOL: proving independence of axioms

I'm given the following axioms:

1. $$\forall x \forall y \forall z(xQy \wedge yQz \rightarrow xQz)$$

2. $$\forall x \forall y \forall z [(x\ast y)\ast z Qx \ast (y \ast z)]$$

3. $$\forall x \forall y \forall z (xQy \rightarrow x \ast zQz \ast y)$$

4. $$\forall x \forall y (\neg xQy \rightarrow (\exists z) (x Q y \ast z \wedge y \ast z Q x))$$

5. $$\forall x \forall y (\neg x \ast y Q x)$$

And I'm trying to prove that the second axiom $$\forall x \forall y \forall z [(x\ast y)\ast z Qx \ast (y \ast z)]$$ is independent of the rest.

In my attempt, I use the following interpretation:

Domain: $$\mathbb{R}^+$$

$$xQy \leftrightarrow x\leq y$$

$$x \ast y \leftrightarrow xy+1$$

So the axioms can be rewritten as:

1. $$\forall x \forall y \forall z(x \leq y \wedge y \leq z \rightarrow x \leq z)$$

2. $$\forall x \forall y \forall z [(xy+1)z+1 \leq x(yz+1)+1]$$

3. $$\forall x \forall y \forall z (x \leq y \rightarrow xz+1 \leq zy+1)$$

4. $$\forall x \forall y (\neg x \leq y \rightarrow (\exists z) (x \leq yz+1 \wedge yz+1 \leq x))$$

5. $$\forall x \forall y [\neg(xy+1 \leq x)]$$

Note that axiom 2 can be rewritten as $$\forall x \forall y \forall z [xyz+z+1 \leq xyz+x+1]$$ Which is false because the inequality doesn't hold for, say, z=y=2 and x = 1.

The next step would be to show that the other axioms are true in this interpretation. In general, to show that a formula governed by a universal quantifier is false, one would only need to provide a counter example. But what steps do I need to take in order to show that such a formula is true? For example, it's quite intuitive that the first axiom: $$\forall x \forall y \forall z(x \leq y \wedge y \leq z \rightarrow x \leq z)$$ is true in this interpretation. But will I have to prove in some systematic way that $$(x \leq y \wedge y \leq z \rightarrow x \leq z)$$ is true for all x, y, and z? How do I formally argue that it's true in this interpretation?

This question may sound silly but appreciate any help or feedback :)

Second, unfortunately your interpretation does not make axiom 5 true: pick $$x=10$$ and $$y=0.1$$ for a counterexample.