Fitch proof (constructing formal proofs) Good official afternoon community, 
I am trying to construct a formal proof for the following argument: 
(P → (Q → R)) ↔ ((P ∧ Q) → R)
I have no idea how to start. This is the outline I put for myself to follow (one of my trials). 
Any help is appreciated. 
Thank you 
Mark
 A: $\vdash (p\to (q\to r)\leftrightarrow ((p\wedge q)\to r)$
For each conditional subproof for this biconditional proof, first assume the antecendant, and further assume what you need to set up for the final conditional introduction.   The argument between should suggest itself. 
$\begin{array}{r|ll}
1. &\quad p\to (q\to r) & \text{Assume}
\\ 2. & \qquad p\wedge q & \text{Assume}
\\ 3. & \qquad \vdots\phantom{p} & \vdots\phantom{2, \text{Conjunction Elimination}}
\\ 4. & \qquad \vdots\phantom{q\to r} & \vdots\phantom{1,3,\text{Conditional Elimination}}
\\ 5. & \qquad \vdots\phantom{q} & \vdots\phantom{2,\text{Conjunction Elimination}}
\\ 6. & \qquad r & \vdots\phantom{4,5,\text{Conditional Elimination}}
\\ 7. & \quad (p\wedge q)\to r & 2,6,\text{Conditional Introduction}
\\ 8. & (p\to(q\to r))\to ((p\wedge q)\to r) & 1,7,\text{Conditional Introduction}
\\[2ex] 7. & \quad (p\wedge q)\to r & \text{Assume}
\\ 8. & \qquad p & \text{Assume}
\\ 9. & \qquad\quad q &\text{Assume}
\\ 10. & \qquad\quad \vdots\phantom{p\wedge q} & \vdots\phantom{9,10,\text{Conjunction Introduction}}
\\ 11. & \qquad\quad r & \vdots\phantom{7,10, \text{Conditional Elimination}}
\\ 12. & \qquad q\to r & 9,11,\text{Conditional Introduction}
\\ 13. & \quad p\to(q\to r) & 8,12,\text{Conditional Introduction}
\\ 14. & ((p\wedge q)\to r)\to(p\to(q\to r)) & 7,13,\text{Conditional Introduction}
\\[2ex] 15. & (p\to(q\to r))\mathop{\leftrightarrow} ((p\wedge q)\to r) & 8,14,\text{Biconditional Introduction}
\end{array}$
A: That's not a good set-up at all.  Since you need to prove a biconditional, look at the rule of $\leftrightarrow$ Intro, and you'll see that you need two subproofs: one going from left to right, and the other one going from right to left.
Here's a trick:
Focus on the desired line and choose the rule through which you intend to derive that line, which is of course $\leftrightarrow$ Intro. Then go to the Proof Menu and select 'Add Support Steps':

Magic will happen! :)
