# Follow the method used in the proof to find a conjunctive form equivalent to $P\to(Q\to(R\lor\lnot P))$

The proof (Theorem 2.5) refers to an induction proof that establishes that all formulae composed of $\{\to, \iff, \lnot, \land, \lor\}$ is logically equivalent to a conjunctive (CF) and disjunctive form (DF). It will be attached at the end of this question.

By CF it means the following: The question also asked for a disjunctive form, which is very similar to CF; but I think I got it, which is $\lnot P\lor(\lnot Q\lor(R\lor\lnot P)$ (or ignore the brackets).

But when it comes to CF I am stuck. So I applied various equivalences to change it to $\lnot P\lor \lnot(Q\land(\lnot R\land P))$ , but this is as close as I can get.

In a nutshell, this seems to be the form I have: $\bigwedge_{i=1}^n\theta_i\lor\lnot \bigwedge_{j=1}^m \psi_j$, where the LHS equals to $\lnot P$ and the RHS equals to $\lnot(Q\land(\lnot R\land P))$. If I can get rid of the negation on the RHS and get the RHS into just a chain of conjunction, I can use 2.41c (as attached below) to get what I need. But I just couldn't get rid of the negation.

Could anyone please help? I have been trying on my own for weeks and I still can't crack it. Thank you so much!

The formula $P→(Q→(R∨¬P))$ is equivalent to: $¬P∨(¬Q∨(R∨¬P))$, that - using distributivity and associativity - is equivalent to:
$(¬P∨¬P)∨(¬Q∨R) \equiv (¬P∨¬Q∨R)$,
It has the form $\bigwedge_{i=1}^1 (\bigvee_{j=1}^3 q_{i,j})$.
In this case $i=1$, because we have only one conjunct.
• Thank you so much! Now it all makes sense because I kept thinking for the last few weeks, that there is simply no way of forcing a formula with a $\land$ as the main operator into one with a $\lor$ (vice versa) as the main operator. But with this it all makes sense. Your answer is very clear and concise. – Daniel Mak Dec 28 '17 at 15:06