
Abstract Algebra Abstract Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 12th, 2014, 08:00 PM  #1 
Newbie Joined: Apr 2014 From: australia Posts: 17 Thanks: 1  propositional calculus
Hi, I'm having trouble proving the following sequent using the rules of deduction in the Propositional Calculus. (P âˆ¨Q)âˆ§(P âˆ¨R) âŠ¢ P âˆ¨(Qâˆ§R) any help would be appreciated. thanks 
April 13th, 2014, 12:57 AM  #2 
Senior Member Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 
Are the two formulas separated by the turnstile? My browser shows just a space between them. There are many variants of rules of deduction in the propositional calculus, and none of them is the default one. Which one are you using? 
April 13th, 2014, 02:55 AM  #3 
Newbie Joined: Apr 2014 From: australia Posts: 17 Thanks: 1 
Yes they are separated by a syntactic turnstile (P âˆ¨Q)âˆ§(P âˆ¨R) âŠ¢ P âˆ¨(Qâˆ§R) I'm thinking to use VElimination, but i'm having trouble reaching the conclusion. 
April 13th, 2014, 06:51 AM  #4  
Senior Member Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108  Quote:
Judging by the name "velimination", you are working with natural deduction. The outline of the derivation is as follows. Derive P \/ Q and P \/ R by /\elimination. Perform a \/elimination on P \/ Q. If P, then P \/ (Q /\ R), so the conclusion is proved. Otherwise, you have an assumption Q. Perform \/elimination on P \/ R. Again, if P, then P \/ (Q /\ R). Otherwise, you have R, and together with Q it gives Q /\ R and P \/ (Q /\ R).  
April 14th, 2014, 07:45 PM  #5 
Newbie Joined: Apr 2014 From: australia Posts: 17 Thanks: 1 
So this is what I have so far 1 (1) (P V Q) & (P V R) A 1 (2) P V Q 1 &E 3 (3) P A 3 (4) P V (Q & R) 3 VI 1 (5) P V R 1 &E 6 (6) Q A 7 (7) R A 6,7 ( Q & R 6, 7 &I 3,6,7 (9) P V (Q & R) 3, 8 VI then Im stuck. . 
April 15th, 2014, 01:23 AM  #6  
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  Quote:
$$\begin{array}{rll} 1. & (P\vee Q)\,\&\,(P\vee R) & \mathrm{(A)} \\ 2. & P\vee Q & \mathrm{(1\ \&E)} \\ 3. & P & \mathrm{(2\ A)} \\ 4. & P\vee(Q\,\&\,R) & \mathrm{(3\ \vee I)} \\ 5. & Q & \mathrm{(2\ A)} \\ 6. & P\vee R & \mathrm{(1\ \&E)} \\ 7. & P & \mathrm{(6\ A)} \\ 8. & P\vee(Q\,\&\,R) & \mathrm{(7\ \vee I)} \\ 9. & R & \mathrm{(6\ A)} \\10. & Q\,\&\,R & \mathrm{(5,9\ \&I)} \\11. & P\vee(Q\,\&\,R) & \mathrm{(10\ \vee I)} \end{array}$$ There are a number of ways to approach the study of propositional calculus, each with its own set of axioms. Another way to approach the problem would be to show that the premise and the negation of the conclusion together form an inconsistent set.  
April 15th, 2014, 07:28 AM  #7 
Senior Member Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 
The derivation in post #5 looks like a derivation in Fitchstyle calculus, or flag notation, which is a form of natural deduction. I'll just add the necessary indentation and disjunction elimination rules to post #6. Code: 1. (P \/ Q) /\ (P \/ R) Assumption 2. P \/ Q 1, /\E 3. P Assumption 4. P \/ (Q /\ R) 3, \/I 5. Q Assumption 6. P \/ R 1, /\E 7. P Assumption 8. P \/ (Q /\ R) 7, \/I 9. R Assumption 10. Q /\ R 5, 9, /\I 11. P \/ (Q /\ R) 10, \/I 12. P \/ (Q /\ R) 6, 78, 911, \/E 13. P \/ (Q /\ R) 2, 34, 512, \/E 
April 16th, 2014, 03:57 AM  #8 
Newbie Joined: Apr 2014 From: australia Posts: 17 Thanks: 1 
thank you for your assistance


Tags 
calculus, propositional 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
introduction propositional calculus  lun123  Applied Math  1  October 21st, 2013 02:28 PM 
Propositional calculus why my proof isn't right?  zcohen  Applied Math  2  April 26th, 2012 05:35 PM 
Propositional Logic  summer90  Applied Math  2  January 27th, 2010 09:26 AM 
Verifying using propositional equivalences  HELP  patel_ankz  Applied Math  1  November 8th, 2008 06:43 PM 
A Propositional Calculus Question  Christi123  Applied Math  0  March 25th, 2008 11:39 PM 