MHB Getting Nowhere with a Proof Question: Help Needed

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The discussion centers around a challenging proof question involving logical equivalences. The original poster is struggling to simplify the expression (¬(Q⇒¬P) ∧ ¬((Q∧¬R)⇒¬P )) ⇔ ¬(R ∨ (P ⇒¬Q)). Participants suggest using logical identities such as ¬(Q ⇒ ¬P) being equivalent to (Q ∧ P) and applying De Morgan's laws. There is some confusion regarding the exact nature of the proof required, indicating a need for clarification on the question itself. Overall, the thread highlights the complexities of proving logical equivalences and the importance of understanding foundational concepts in logic.
Leanna
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I'm stuck on this proof question:
(¬(Q⇒¬P) ∧ ¬((Q∧¬R)⇒¬P )) ⇔ ¬(R ∨ (P ⇒¬Q))

I've tried to get rid of the negation and implications but I keep going in circles and I'm getting nowhere near to the equivalence required. I would appreciative if anyone can help me solve this because it's really been doing my head in :/
 
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If you see question marks that's the negation
 
Hi Leanna,

I'm not sure what kind of proof structure is required, but you can obtain the results if you use the following:
$\lnot (Q \implies \lnot P) \iff (Q \land P)$, De Morgan's, and the fact that $Q \land Q \land P \iff Q \land P$.
 
Leanna said:
I'm stuck on this proof question:
(¬(Q⇒¬P) ∧ ¬((Q∧¬R)⇒¬P )) ⇔ ¬(R ∨ (P ⇒¬Q))
What exactly is the question? What you have written is a formula.
 
Greetings, I am studying probability theory [non-measure theory] from a textbook. I stumbled to the topic stating that Cauchy Distribution has no moments. It was not proved, and I tried working it via direct calculation of the improper integral of E[X^n] for the case n=1. Anyhow, I wanted to generalize this without success. I stumbled upon this thread here: https://www.physicsforums.com/threads/how-to-prove-the-cauchy-distribution-has-no-moments.992416/ I really enjoyed the proof...

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