I am asked to use fitch to prove this statement (P → (Q → R)) ↔ ((P ∧ Q) → R)

Question

I am asked to use fitch to prove this statement (P → (Q → R)) ↔ ((P ∧ Q) → R)

This is what i've tried:

Answer 1

First of all, always make sure that the last line of your proof is the actual goal you need to prove. In your attempt, you have a conditional as your last line (main connective is $\to$), but the goal of this problem is to prove a bi-conditional (main connective is $\leftrightarrow$). So, first put the appropriate statement there:

OK, now, since you are trying to prove a bi-conditional, set it up as a biconditional proof. i.e. choose the rule for $\leftrightarrow$ Intro:

Now, rather than typing out all the lines you need, there is a really handy feature in Fitch: Go to the Proof menu, and select 'Add Support Steps':

Magic happens!

Of course, you could have typed this all out yourself as well, and in fact, as a beginner, I would urge you to in fact not constantly rely on 'Add Support Steps', but work these out by yourself, until you got the pattern down, and then use the handy-dandy 'Add Support Steps' feature to save some typing.

Anyway, you now have the basic set-up to prove the biconditional: complete these two subproofs, and you're done. Now, you of course effectively tried the very first subproof, and unfortunately you got stuck. This is because you didn't set it up right.... set-up is key!!

OK, the goal of the first subproof is to prove $(P \land Q) \to R$, and so you need to do a conditional proof, where you assume the 'if' part ($P \land Q$), and try and end up with the 'then' part ($R$):

Now, can you try and finish it yourself from here?

Answer 2

This is what i've tried:

Why? What was your motivation for those assumptions?

Natural deduction proofs don't work by randomly making assumptions hoping for magic to happen. The goal practically outlines its proof. That's why the checker can automate the setup for you. However, you can do it for yourself.

Since you want to prove $(p\to(q\to r))\leftrightarrow((p\land q)\to r)$ you should:

• Assume $(p\to(q\to r))$ and $(p\land q)$ aiming to derive $r$, and do that!
  • $\therefore (p\to(q\to r))~\to~((p\land q)~\to~ r)$
• Assume $((p\land q)\to r)$, $p$, and $q$ aiming to derive $r$, and do that!
  • $\therefore ((p\land q)\to r)~\to~ (p~\to~ (q~\to~ r))$
• $\therefore (p\to(q\to r))\leftrightarrow((p\land q)\to r)$
• Done.

Well, except for the actual derivations, but these are not obscure.

$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline #2\end{array}}\fitch{}{\fitch{p\to(q\to r)}{\fitch{p\land q}{~~\vdots\\r}\\(p\land q)\to r}\\(p\to(q\to r))\to((p\land q)\to r)\\\fitch{(p\land q)\to r}{\fitch{p}{\fitch{q}{~~\vdots\\r}\\q\to r}\\p\to(q\to r)}\\((p\land q)\to r)\to(p\to (q\to r))}$$