Proving $(ab)^{-1} = a^{-1}b^{-1}$, if $a,b\ne 0$

Question

Having only these axioms:

• add associativity.
• add identity.
• add inverse.
• add commutative.
• mul associativity.
• mul identity.
• mul inverse.
• mul commutative.
• distributive.

Prove that $(ab)^{-1} = a^{-1}b^{-1}$, if $a,b\ne 0$

My attempt 1 (edit: false)

Using [mul identity]: $$\begin{split} (ab)^{-1} &=(ab)^{-1}\times 1 \times 1\\ \end{split}$$

Using [mul associativity]: $$\begin{split} (ab)^{-1}\times 1 \times 1 &= (a^{-1}1) \times (b^{-1}1)\\ \end{split}$$

Using [mul identity]: $$\begin{split} (a^{-1}1) \times (b^{-1}1) &= a^{-1} b^{-1}\\ \end{split}$$

$\blacksquare$

My attempt 2 (edit: false)

Using [mul associativity]: $$\begin{split} (ab)^{-1} &= a^{-1} b^{-1}\\ \end{split}$$

$\blacksquare$

My attempt 3 (I think I nailed it here)

If $a$ and $b$ are numbers, then their product, $(ab)$, is a number too.

By [mul inverse], we know that: $$(ab)(ab)^{-1}=1$$

We also know that the equality holds if we multiply both sides by the same numbers:

$$(ab)(ab)^{-1} a^{-1} b^{-1}=(1) a^{-1} b^{-1}$$

By [mul associativity] we know: $$(aa^{-1}) (bb^{-1}) (ab)^{-1} =(1) a^{-1} b^{-1}$$

By [mul inverse] we know:

$$(1) (1) (ab)^{-1} =(1) a^{-1} b^{-1}$$

By [mul identity] we know:

$$(ab)^{-1} = a^{-1} b^{-1}$$

$\blacksquare$

Questions

My goal is to achieve maximum rigor based on the axioms above.

1. What are the mistakes in my 1st attempt?
2. What are the mistakes in my 2nd attempt?
3. What is the best way of proving it?

Answer 1

My proof.

\begin{align*} &\quad (ab)(a^{-1}b^{-1})\\ &= (aba^{-1})b^{-1}\tag{mul asso}\\ &= (aa^{-1}b)b^{-1} \tag {mul comm}\\ &= (1b)b^{-1} \tag {mul inv}\\ &= 1(bb^{-1}) \tag {mul asso}\\ &= 1\cdot 1 \tag {mul inv}\\ &= 1. \tag {mul identity} \end{align*} By the definition of multiplicative inverse, $(ab)^{-1} = a^{-1}b^{-1}$.