Fitch natural deduction proof of $(P \to Q) \land R, (P \land R) \to S, \neg S \vdash P \to (Q \land R)$

Question

I just got stuck and need help seeing what other steps I can take.

Answer 1

This answer expands on lemontree's comment in case that wasn't clear.

Instead of making an assumption on line 5 of $P \to Q$, derive $P \to Q$ using conjunction elimination ($\land$ Elim) referencing line 1 which is the premise $(P \to Q) \land R$.

Lines 6, 7, and 8 should stay the same.

On line 9 you can discharge the assumption $P$ that you made on line 4 using conditional introduction ($\to$ Intro) referencing lines 4-8. This will give you the desired result $P \to (Q \land R)$.

That should complete the proof.

Answer 2

This attempt is not in Fitch style ( I believe) . It only tries to show that, with some equivalence rules at hand, the conclusion can be reached without using conditional proof.