Duality Principle in Boolean Algebra - Why do I alway get !F instead of F?

Question

I have the function:

F = !(a && d || b || c)

Now i apply the duality principle and exchange all * with +

Fd = !((a || d) && b && c)

Which is !F and not as I expected F.

Another principle says that I get the same result by complementing all variables in F and the expression itself:

fd = (!a && !d || !b || !c)

Again I get the expression !F and not the expected F.

Why is this the case?

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To answer the comments, is this the case because I can imagine 0 and 1 instead of F and have to negate that as well?

Answer 1

It seems that there is a little bit of terminological confusion on this issue.

See Steven Givant & Paul Halmos, Introduction to Boolean algebras (2009), Ch.4 : The Principle of Duality, page 4 :

Every Boolean polynomial has a dual: it is defined to be the polynomial that results from interchanging $0$ and $1$, and at the same time interchanging ∧ and ∨.

A slight misunderstanding can arise about the meaning of duality, and often does. It is well worthwhile to clear it up once and for all, especially since the clarification is quite amusing in its own right. If an experienced Boolean algebraist is asked for the dual of a Boolean polynomial, such as say $p∨q$, his answer might be $p∧q$ one day and $\lnot p ∧ \lnot q$ another day; the answer $\lnot p ∨ \lnot q$ is less likely but not impossible.

What is needed here is some careful terminological distinction. Let us restrict attention to the completely typical case of a polynomial $f(p, q)$ in two variables. The complement of $f(p, q)$ is by definition $\lnot (f(p, q))$; the dual of $f(p, q)$ is $\lnot f (\lnot p , \lnot q)$; the contradual of $f(p, q)$ is $f(\lnot p , \lnot q)$.

The polynomial

$$(1) \quad p ∧ (q ∨ (\lnot p ∧ 0))$$

may serve as an example. Its complement is formed by applying the operation $\lnot$ to the entire expression, and then simplifying the result with the help of the De Morgan laws and the double complement [or double negation] law :

$$\lnot p ∨ (\lnot q ∧ (p ∨ 1))$$

The contradual is formed by replacing $p$ and $q$ in (1) with their complements, and then simplifying:

$$\lnot p ∧ (\lnot q ∨ (p ∧ 0))$$

The dual is the complement of the contradual, appropriately simplified:

$$p ∨ (q ∧ (\lnot p ∨ 1))$$