# Transform to CNF (conjunctive normal form)

I am trying to convert the following expression to CNF (conjunctive normal form):

$$\left(A\Rightarrow B\right)\Rightarrow\left(A\Rightarrow C\right)$$

As my first steps I am removing the implications like so:

$$\left(\lnot A ∨ B\right)\Rightarrow\left(\lnot A ∨ C\right)$$

$$\lnot\left(\lnot A ∨ B\right)∨\left(\lnot A ∨ C\right)$$

Removing the negation by applying it to the parentheses:

$$\left(A\land\lnot B\right)\vee\left(\lnot A\vee C\right)$$

Until here everything is clear for me. This next step where (I think) the distributive law is used is not clear to me:

$$\left(A\vee\lnot A \vee C\right)\land\left(\lnot B \vee\left(\lnot A \vee C\right)\right)$$

(The answer the teacher gave us)

Those steps are all correct, and the last step uses the distributed law $$(P \land Q) \lor R \equiv (P \lor R) \land (Q \lor R)$$
What your teacher provided is CNF but can be further simplified because $$A\lor \lnot A$$ is always true, yielding $$\lnot B \lor \lnot A \lor C$$.
• Yes, the $R$ can be any logical expression. In this case, it is $\lnot A \lor C$. Sep 27 at 19:24