DNF of B∧(A∧B∧(A ∨ B ∨ (B∧C)))

Question

if you write

$B∧(A∧B∧(A ∨ B ∨ (B∧C)))$ like

$B ∧ A ∧ B ∧ (A ∨ B ∨ (B ∧ C))$ you can see better the idempotence of $∧ (B∧B = B)$ so:

$$A ∧ B ∧ (A ∨ B ∨ (B ∧ C))$$

My idea was to use distribution inside the parenthesis:

$((A ∨ B) ∨ (B ∧ C))$ equivalent to $(B ∨ A ∨ B) ∧ (C ∨ A ∨ B)$

where you can use idempotence of $∨$ to get:

$$A ∧ B ∧ ((A∨B) ∧ (C ∨ A ∨ B))$$

How do I get from here to

$$(A ∧ B) ∨ (B ∧ C ∧ A) $$

Which rules do I use?

I understand that, when I distribute over ∧ that the conjunction inside the parenthesis changes to a disjunction, but does it matter, which atomic sentence I distribute over which Literal first.

I also know, that after the distribution I can use the idempotence of junctions but I'm again confused about the "correct or most efficient" way to do so.

Thanks!

Answer 1

You can do everything here with:

Absorption

$A \land (A \lor B) = A$

$A \lor (A \land B) = A$

Using Absorption twice you can do:

$A \land B \land (A\lor B \lor (B \land C))= A\land B \land (A\lor B) = A \land B$

In fact, you could have the term $(A\lor B \lor (B \land C)$ immediately be absorbed to do this in a single step:

$A \land B \land (A\lor B \lor (B \land C))= A \land B$

Note that $A \land B$ is already in DNF.

But if the goal is to get to $(A \land B) \lor (B \land C \land A)$, note that we can use Absorption to introduce terms as well:

$A\land B = (A \land B) \lor (A \land B \land C)$

and now it’s a simple Commutation to get to your goal.