Full derivation of ∼(P→Q)→Q ∴ P→Q

Computer Science
Tutor: None Selected Time limit: 1 Day

 ∼(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

Are you studying on the go? Check out our FREE app and post questions on the fly!
Download on the
App Store
...
Jul 6th, 2015
...
Jul 6th, 2015
Dec 5th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer