Transform a logical function into conjunctive form

Question

For this function, i need to find the clause form for $A\Rightarrow B$ where $$A \equiv (A\land B\land \sim C) \lor (\sim A \land C)$$ and $$B\equiv (\sim A \land \sim B\land \sim C)$$

I understand that in order to find the clause form I need to put this function in conjunctive form and then all the sub-function that formed the conjunction formula are my elements of the clause form. I can't transform this formula in conjunctive form.

In the solution, I found that one transformation leads to this formula: $$A \Rightarrow B \equiv (\sim A \lor \sim B\lor C \lor \sim C)\land (\sim A \lor A \lor \sim B\lor \sim C)$$

I tried every step and properties I know but cannot find a way to prove the equivalence of these functions. What are the necessary steps to arrive at this form?

Answer 1

Let :

$\varphi := (A∧B∧∼C)∨(∼A∧C)$

and :

$\psi := (∼A∧∼B∧∼C)$.

We have to consider $\varphi \to \psi$.

We can apply the equivalence between $p \to q$ and $\lnot p \lor q$, followed by De Morgan's laws to get :

$$\lnot \varphi \lor \psi := \lnot [(A∧B∧∼C)∨(∼A∧C)] ∨ (∼A∧∼B∧∼C) \equiv$$

$$\equiv [(∼A∨∼B∨C)∧(A∨∼C)] ∨ (∼A∧∼B∧∼C)$$

By Distributivity we get :

$$[(∼A∨∼B∨C) ∨ (∼A∧∼B∧∼C)] ∧ [(A∨∼C) ∨(∼A∧∼B∧∼C)]$$

Consider now the LHS and apply distributivity; thus:

$$(∼A∨∼B∨C) ∨ (∼A∧∼B∧∼C) \equiv [(∼A∨∼B∨C) ∨ ∼A] ∧ [(∼A∨∼B∨C) ∨ ∼B] ∧ [(∼A∨∼B∨C) ∨ ∼C]$$

But $C ∨ ∼C \equiv T$ and $\alpha ∨ T \equiv T$ and $\beta ∧ T \equiv \beta$.

Thus, for the LHS we have :

$$[(∼A∨∼B∨C) ∨ ∼A] ∧ [(∼A∨∼B∨C) ∨ ∼B] \equiv$$

$$(∼A∨∼B∨C) ∧ (∼A∨∼B∨C)$$

Now consider the RHS and apply distributivity again:

$$[(A∨∼C) ∨ (∼A∧∼B∧∼C)] \equiv$$

$$[ (A∨∼C) ∨ ∼A] ∧ [(A∨∼C) ∨ ∼B] ∧ [(A∨∼C) ∨ ∼C] \equiv$$

$$(A∨∼C ∨ ∼B) ∧ (A∨∼C)$$

Now we collect then together to have :

$$(∼A∨∼B∨C) ∧ (∼A∨∼B∨C) ∧ (A∨∼C ∨ ∼B) ∧ (A∨∼C)$$