# Show that $\alpha$ is independent of the set of axioms (First Order Logic)

Let L be the group language with a constant "e" and functional symbol ".".

Considering the following set $\Gamma$ of axioms:

$\forall$$x((x.e = x) \land (e.x=x)) \forall$$x\exists$$y(x.y=e) \forall$$x$$\forall$$y$$\forall$$z(x.(y.z) = (x.y).z)$

Let $\alpha$ be the sentence $\forall$$x(x.x=e). Show that \alpha is independent of the axioms, determining a model to \Gamma \cup {\alpha} and a model to \Gamma \cup {¬\alpha}. A model to \Gamma \cup {¬\alpha} could be Q = {\mathbb{Q}+$$,1, .$} (with . as the usual multiplication), but what could be a model to $\Gamma$ $\cup$ {$\alpha$}?

The listed axioms are the axioms defining a group, the sentence $\alpha$ states that every element is its own inverse and a model would be the trivial group.