writing a formal proof If B is a statement form involving only negation, conjunction and disjunction, and B' results from B by replacing each conjunction by a disjunction and each disjunction by a conjunction,  show that B is a tautology if and only if non-B' is a tautology.
It is clear that if non-B' is a tautology then B is also a tautology.That is because B' is obtained from B by switching the connectors so by applying the negation to B' we obtain B. 
I just don't know how to write a formal proof.Any help would be appreciated.
 A: We need a couple of propositions before proving the statement.
First, define another transform $A^{\neg}$ of propositional formulas $A$: $A^{\neg}$ is the results of negating all propositional variables in $A$.
This transform leaves all connectives unchanged — it distributes over $\neg, \land, \lor$. Note that the two transforms commute: $A'^{\neg} = {A^{\neg}}'$.
For example, if $B = (p\lor \neg p)$ then $B' = (p\land \neg p)$ and $B'^{\neg} = (\neg p\land \neg\neg p)$.
Proposition 1: $B$ is equivalent to $\neg B'^{\neg}$. 
We'll use $\equiv$ to denote logical equivalence. Proof is by induction on the complexity of $B$.
Base case: $B$ is some propositional variable $p$. Then $B' = p$, and $B'^{\neg} = \neg p$, so $B = p \equiv \neg\neg p = \neg B'^{\neg}$.
[$B = \neg A$]:  $B'$ is $\neg A'$, and $B'^{\neg}$ is $\neg A'^{\neg}$. By induction hypothesis (IH), $A\equiv \neg A'^{\neg}$, so clearly $B = \neg A\equiv \neg\neg A'^{\neg} = \neg B'^{\neg}$.
[$B = (C\lor D)]$: $B' = (C'\land D')$, and $B'^{\neg} = (C'^{\neg}\land D'^{\neg})$.
By De Morgan, $\neg B'^{\neg} \equiv (\neg C'^{\neg} \lor \neg D'^{\neg})$. By IH, $C\equiv \neg C'^{\neg}$ and $D\equiv \neg D'^{\neg}$, so $\neg B'^{\neg} \equiv (C \lor D) = B$.
[$B = (C\land D)]$: Similar to the previous case, using the other De Morgan law.
Proposition 2: if $A$ is a tautology iff $A^{\neg}$ is a tautology.
Given any assignment of truth values to propositional variables $v\colon Variables\to \{0,1\}$, where $0$ and $1$ represent False and True respectively, let $v^{\neg}$ be the assignment that "flips" $v$: $$v^{\neg}(p) = 1 - v(p).
$$
Obviously this operation on assignments is self-inverse:
$$v^{{\neg}{\neg}} = v, \tag{SI}
$$
and a bijection of assignments.
For any formula $A$, let $A[v]$ be the truth value of $A$ under the assignment $v$. Clearly, 
$$A^{\neg}[v] = A[v^{\neg}].\tag{*}
$$
By definition, $A$ is a tautology iff for all assignments $v$, $A[v] = 1$. By (SI), $A$ is a tautology iff for all assignments $v$, $A[v^{\neg}] = 1$. By (*), this holds iff for all assignments $v$, $A^{\neg}[v] = 1$, which is true iff $A^{\neg}$ is a tautology.
Now we can prove the statement:

$A$ is a tautology iff $\neg A'$ is a tautology.

By Prop. 1, $A\equiv \neg (A'^{\neg})$, which therefore is a tautology iff $A$ is. But $\neg (A'^{\neg}) = (\neg A')^{\neg}$, which by Prop. 2 is a tautology iff $\neg A'$ is. This yields the result.
