Theorem

Let G be any group with operation $\ast$ .

\forall \;g\in G:[g^{-1}]^{-1}=g

In Group G, inverse of inverse of any element g is g.

Proof

0. Choose ${\color {OliveGreen}g}\in G$
1. $\exists \;{\color {BrickRed}g^{-1}}\in G:{\color {OliveGreen}g}\ast {\color {BrickRed}g^{-1}}={\color {BrickRed}g^{-1}}\ast {\color {OliveGreen}g}=e_{G}$	definition of inverse of g in G (usage 1,3)
2. ${\color {OliveGreen}g}\ast {\color {BrickRed}a}={\color {BrickRed}a}\ast {\color {OliveGreen}g}=e_{G}$	let a = g⁻¹
3. ${\color {BrickRed}a}\ast {\color {OliveGreen}g}={\color {OliveGreen}g}\ast {\color {BrickRed}a}=e_{G}$
4. $[{\color {BrickRed}a}]^{-1}={\color {OliveGreen}g}$	definition of inverse of a in G (usage 2)
5. $[{\color {BrickRed}g^{-1}}]^{-1}={\color {OliveGreen}g}$	as a = g⁻¹