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

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

• I added a screenshot – Jonjon May 5 at 18:05
• Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – Brian May 5 at 18:11
• The error messages often give you useful information. What do they say? – lemontree May 5 at 18:56
• If $P \to (Q \land R)$ is the conclusion you need to show, then you don't need premises 2. and 3. at all and lines 9-17 are not necessary. In any case, you can get $(P \to Q)$ directly from 1. on the level of the subproof with assumption P by use of $\land E$, instead of opening a new subproof and making $P \to Q$ only an assumption. With that, lines 1-8 will lead you directly to the conclusion in line 18. – lemontree May 5 at 19:04
• That being said, please specify explicitly in the description of your question what the given premises and the conclusion to arrive at are (like I edited into the question title), so we know what you need to prove even if your attempt is incomplete or the image is not readable. If my edit is incorrect, please fix it. – lemontree May 5 at 19:06

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.

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.