Full derivation of ∼(P→Q)→Q ∴ P→Q

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 ∼(P→Q)→Q ∴ P→Q

with the rules used

Jul 6th, 2015

Thank you for the opportunity to help you with your question!

Here is my solution in .doc file.
solution.docx


The formula

∼(P→Q)→Q ∴ P→Q

 is derived.

Please let me know if you need any clarification. I'm always happy to answer your questions. *And finally, please don't forget to give me a review in the end. I would be highly appreciate that.
Jul 6th, 2015

is it possible to add the rules ? and have it i the format of : show , ass cd, etc..

Jul 6th, 2015

I use the rule ~(X → Y) ∴ ~X v Y for all logical expressions X, Y to convert ∼(P→Q)→Q to only "AND" and "OR" form, which then allows me to add. The final expression turns out to be ~P v Q which is precisely P → Q.

Hope this answer your question.

Jul 6th, 2015

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Jul 6th, 2015
Jun 27th, 2017
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