Answer:
Case ¬ϕ: the equivalence holds, since f(ϕ) ⊼ f(ϕ) is false if and only if f(ϕ) is true, thus it negates f(ϕ), which is equivalent to ϕ. In the truth table below, where we get the valuation in the third column from the induction hypothesis (IH), we see that the columns marked with arrows agree, and thus the formulas are semantically equivalent.
The thing I don't understand is, where did they even use the induction hypothesis in the proof for "and", and on the negation part I don't really understand how the induction hypothesis is helping us in any way?
To clarify our IH is $f(\phi) = \phi$ if I understood it correctly, then my question is what is this fact telling us? In the truth table everything is already clear to me without induction, tbh I don't even see the point or the use of it in this case.
Long comemnt
$\phi$ is a formula and the IH for e.g. the negation case is used in the assumption that $f(\phi)$ and $\phi$ are equivalent.
The proof by induction has the following structure, where the "counter" of the induction is the number of conenctives:
Base case: no connectives, i.e. $\phi$ is an atom, like $p,q,\ldots$.
In this case $f(\phi)=\phi$ (equal) and the result (equivalence) obviously holds.
The next step is the:
Induction step, and is twofold:
(i) the negation case, i.e. prove that IF $f(\phi)$ is equiv to $\phi$, THEN $f(\lnot \phi)$ is equiv to $\lnot (\phi$); and
(ii) the binary connectives case (all similar), i.e. prove that IF $f(\phi)$ equiv $\phi$ and $f(\psi)$ equiv $\psi$, THEN $f(\phi \square \psi)$ is equiv to $(\phi \square \psi)$, where $\square$ is one of the binary connectives.