Discuss the step that involves the liberation of a proton from the $\sigma$-complex.

(N/A) In the first slow step,an electrophile $(E^+)$ attacks the benzene ring to form a $\sigma$-complex (arenium ion),which is a resonance-stabilized carbocation.
In the second step,which is a fast step,the $\sigma$-complex loses a proton $(H^+)$ to a base (the anion $A^-$ formed from the catalyst).
The reaction is represented as:
$\sigma\text{-complex} + A^- \longrightarrow \text{Substituted product} + H^+A^-$
This step restores the aromaticity of the ring,as the $sp^3$ hybridized carbon atom in the $\sigma$-complex is converted back to an $sp^2$ hybridized carbon atom in the final substituted product.
Examples:
$FeCl_4^- + H^+ \longrightarrow HCl + FeCl_3 + C_6H_5Cl$
$FeBr_4^- + H^+ \longrightarrow HBr + FeBr_3 + C_6H_5Br$
Since this step is fast,it is not the rate-determining step of the reaction.