Showing that $\mathscr B$ is a tautology iff $\lnot \mathscr B'$ is a tautology. 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.

But I think there is a counterexample if  $\mathscr B = A$ , then  $\mathscr B' = A$ .If  $\mathscr B$  is a tautology , then $\lnot  \mathscr B' = \lnot A$ will be a contradictory and this contradicts the thing we want to prove.I am confused with this.Is there a flaw in my reasoning?
 A: As Mauro ALLEGRANZA points out, your counterexample does not work, since if $A$ is a propositional variable, it can not be a tautology: just take a valuation for which it is mapped to $0$.
I will prove a slightly stronger result by induction: for a valuation $\nu$, define $\nu^{*}$ as the valuation such that, for any propositional variable $A$, $\nu^{*}(A)=1$ if, and only if, $\nu(A)=0$. Then I state that
$$\nu(\mathfrak{B})=\nu^{*}(\neg\mathfrak{B}'),$$
for any formula $\mathfrak{B}$.

*

*Suppose, first, that the formula $\mathfrak{B}$ has no connectives between $\vee$ and $\wedge$: since we are considering only formulas with $\vee$, $\wedge$ and $\neg$ as connectives, we must have $\mathfrak{B}=\neg\cdots\neg A$, for a propositional variable $A$ and some number of negations applied to $A$. In that case, $$\mathfrak{B}'=\mathfrak{B}\quad\text{and so}\quad\neg\mathfrak{B}'=\neg\mathfrak{B},$$ and one easily sees (apply induction in the number of negations of $\mathfrak{B}$ if you must) that $\nu(\mathfrak{B})=\nu^{*}(\neg\mathfrak{B}')$.


*Suppose now that the result holds for formulas with, at most, $n-1$ ($n\geq 1$) connectives in $\{\vee, \wedge\}$, and take $\mathfrak{B}$ with $n$ of these connectives; we can write
$$\mathfrak{B}=\neg\cdots\neg(\mathfrak{C}\#\mathfrak{D})=\neg^{m}(\mathfrak{C}\#\mathfrak{D}),$$
for some $\#\in\{\vee, \wedge\}$ and some number $m$ of negations (possibly zero) applied to $\mathfrak{C}\#\mathfrak{D}$. Then we have that $\mathfrak{B}'=\neg^{m}(\mathfrak{C}'\#'\mathfrak{D}')$, where $\vee'=\wedge$ and $\wedge'=\vee$. Since $\mathfrak{C}$ and $\mathfrak{D}$ have fewer connectives among $\{\vee, \wedge\}$ than $\mathfrak{B}$, the induction hypothesis applies to them and then
$$\nu^{*}(\neg\mathfrak{B}')=\nu^{*}\Big(\neg^{m+1}(\mathfrak{C}'\#'\mathfrak{D}')\Big)=\nu^{*}\Big(\neg^{m}(\neg\mathfrak{C}'\#\neg\mathfrak{D}')\Big)=-^{m}\Big(\nu^{*}(\neg\mathfrak{C}')\#\nu^{*}(\neg\mathfrak{D}')\Big)=$$
$$-^{m}\Big(\nu(\mathfrak{C})\#\nu(\mathfrak{D})\Big)=-^{m}\nu(\mathfrak{C}\#\mathfrak{D})=\nu\Big(\neg^{m}(\mathfrak{C}\#\mathfrak{D})\Big)=\nu(\mathfrak{B}),$$
where: by "$-$" I mean the complement on the two-valued Boolean algebra (so $-0=1$ and $-1=0$); and to show $\nu^{*}(\neg^{m+1}(\mathfrak{C}'\#'\mathfrak{D}'))=\nu^{*}(\neg^{m}(\neg\mathfrak{C}'\#\neg\mathfrak{D}'))$ I used De Morgan's law. This finishes our proof.
Finally, you prove the result you want to prove by using this as lemma: if $\mathfrak{B}$ is a tautology, by noticing that $\nu^{**}=\nu$, for any valuation $\nu$ one has $\nu(\neg\mathfrak{B}')=\nu^{*}(\mathfrak{B})$, which equals $1$ since $\mathfrak{B}$ is a tautology.
Reciprocally, if $\neg\mathfrak{B}'$ is a tautology, for any valuation $\nu$ one has $\nu(\mathfrak{B})=\nu^{*}(\neg\mathfrak{B}')$, which is $1$ since $\neg\mathfrak{B}'$ is a tautology.
