Abstract Algebra/Group Theory/Group/Cancellation

=Theorem=
 * Let G be a Group.
 * 1. $$ \forall \; g, a, b \in G: (g \ast a = g \ast b) \rightarrow (a = b) $$
 * 2. $$ \forall \; g, a, b \in G: (a \ast g = b \ast g) \rightarrow (a = b) $$

=Proof=
 * {| style="width: 60%"

|-    |-     |-     |-     |-     |- =Diagrams=
 * 0. Choose $$ {\color{OliveGreen}g}, a, b \in G $$ such that $${\color{OliveGreen}g} \ast a = {\color{OliveGreen}g} \ast b$$
 * 1. $$ {\color{BrickRed}g^{-1}} \in G$$
 * definition of inverse of g in G (usage 1)
 * 2. $$ {\color{BrickRed}g^{-1}} \ast ({\color{OliveGreen}g} \ast a) = {\color{BrickRed}g^{-1}} \ast ({\color{OliveGreen}g} \ast b) $$
 * 0.
 * 3. $$ ({\color{BrickRed}g^{-1}} \ast {\color{OliveGreen}g}) \ast a = ({\color{BrickRed}g^{-1}} \ast {\color{OliveGreen}g}) \ast b $$
 * $\ast$ is associative in G
 * 4. $$ e_G \ast a = e_G \ast b $$
 * g-1 is inverse of g (usage 3)
 * 5. $$ a = b \, $$
 * eG is identity of G (usage 3)
 * }

=Usage=
 * 1) if a, b, x are in the same group, and x*a = x*b, then a = b

=Notice=
 * a, b, and g have to be all in the same group.
 * 1) $$\ast$$ has to be the binary operator of the group.
 * 2) G has to be a group.