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How would I prove this in Fitch?

((P ->R) ^ (Q->notR)) -> (Q->notP)

More generally, what is a strategy I can use to tackle these types of problems in general? I tried working backwards, breaking everything into smaller pieces. I got the point where I had (P->R) and (Q->notR), but I didn't know how to use those once I got there.

  • P →R gives you ~R →~P. Since also Q →~R you get ~P out of P. This gives ~P out of Q by reductio. – Conifold Nov 17 '19 at 9:09
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I used the website https://proofs.openlogicproject.org/ to generate the proof

The good news is that your intuition is correct, and you got steps 1-3 correct. But, from the OP, you struggled with what action to take in step 4.

Honestly, it comes from what you're trying to prove. As a general rule:

If the conclusion you are trying to prove is a material conditional then start by either 1) make a sub-proof starting with the antecedent (Q) and see if you can derive the consequent (~P). That is what I did above.

2) Assume the negation of what you are trying to prove ~(Q->~P). This would have worked but it would require a longer and more complex proof.

3) Keep in mind more advanced proofs may require the use of Implication and De Morgan's laws.

Note: I used the website https://proofs.openlogicproject.org/ to 
format and check my proof. Line 8's justification seems awkward to me, but 
it's the notation used by the site. The logic is that R & ~R (5, 7 ^I) 
violates the PNC and thus leads to an absurdity. Depending on the 
book/professor you learned from, you may be familiar with a different 
style or notation, but what is important is the concept of negation 
introduction (~I) that I use at line 9, which is a key concept in natural logic. 

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