This lemma is what i want to proof (in propositional logic):
$I(B_1\leftrightarrow B_2) \leq I(A[p_i/B_1] \leftrightarrow A[p_i/B_2]) $
This is what i have done:
For $I(B_1\leftrightarrow B_2)=0$ right handside of $\leq$ isn't important so we just do the induction for $I(B_1\leftrightarrow B_2)=1$
\begin{equation} P=Atoms \\ \text{IB}: A\in P \begin{cases} \text{$I(B_1 \leftrightarrow B_2)=1$ }&\quad\text{if $A = p_i$} \\ \text{$I(A \leftrightarrow A)=1$}&\quad\text{if $A \neq p_i$} \end{cases} \\ \text{IS} : 1 \leq I(C[p_i/B_1] \leftrightarrow C[p_i/B_2]) \\\Rightarrow I(C[p_i/B_1] \leftrightarrow C[p_i/B_2]) =1 \\1 \leq I(D[p_i/B_1] \leftrightarrow D[p_i/B_2]) \\\Rightarrow I(D[p_i/B_1] \leftrightarrow D[p_i/B_2]) =1 \\ \vdots \\(\text{how?})\\ \vdots \\ I\Big(\left(C[p_i/B_1] \circ D[p_i/B_1]\right) \leftrightarrow \left(C[p_i/B_2] \circ D[p_i/B_2]\right) \Big)=1 \\ \circ \in \left\{ \land,\lor,\rightarrow \right\} \end{equation}
I dont know how to reach there.
[Update:]
I consider every connectives, but i think my proof is too long , this is what i proof in my whiteboard:
