I have the following logic question about duality:(Introduction to mathmatical Logic by elliot mendelson , exercise $1.30$)
If $\mathscr B$ is a statement form involving only $\lnot$ , $\land$, and $\lor$, and $\mathscr B'$ results from $\mathscr B$ by replacing each $\land$ by $\lor$ and each $\lor$ by $\land$ , show that $\mathscr B$ is a tautology if and only if $\lnot \mathscr B'$ is a tautology.
Now , at first , I had some confusion about the question itself which I resolved with another post of mine. After that , While trying to prove it , I got stuck for a while until I found another post which has the same question If $\phi$ is a tautology then dual $\phi$ is a contradiction.
There , They defined a couple of things:
$1$.In general, for any two-place logical operator $\times$, the dual expression of $\phi \times \psi$ is $\phi^d \times^d \psi^d$, where $\times^d$ is the dual operator of $\times$, which you can find by: $\phi \times^d \psi \Leftrightarrow \neg (\neg \phi \times \neg \psi)$.
$2$.Define the complement statement $\phi'$ of a statement $\phi$ to be that statement that puts a negation in front of every atomic statement in $\phi$. Recursively:
$A' = \neg A$ for any atomic $A$
$(\neg \phi)' = \neg (\phi)'$
$(\phi \times \psi)' = \phi' \times \psi'$
Also the answer states:"you should be able to prove (by induction of course) that $\phi^d\Leftrightarrow \neg \phi'$. And from that, the desired result follows fairly quickly."
I was able to prove that $\phi^d\Leftrightarrow \neg \phi'$ .But I stuck on figuring out how i will use this corollary to prove that " $\phi$ is a tautology iff $\phi^d$ is a contradiction" .Is it something really trivial I am missing out?
Edit: Here is the question again from mendelsons book but I rewrote some symbols in the context of the symbols used in this post so that it is easier to understand,
If $\phi$ is a statement form involving only $\lnot$ , $\land$, and $\lor$, and $\phi^d$ results from $\phi$ by replacing each $\land$ by $\lor$ and each $\lor$ by $\land$ , show that $\phi$ is a tautology if and only if $\lnot \phi^d$ is a tautology.