SOLUTION: (PvQ)&R (R&P)>S (Q&R)>S Therefore, S

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Question 1042390: (PvQ)&R
(R&P)>S
(Q&R)>S
Therefore,
S
Found 3 solutions by solver91311, Edwin McCravy, robertb:
Answer by solver91311(24713) About Me  (Show Source):
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 1.  (P v Q) & R
 2.  R & P -> S
 3.  Q & R -> S        |  S

 4.  P v Q                1  Conjunction Elimination
 5.  R                    1  Conjunction Elimination
|   6.  P                    ACP, Case 1
|   7.  R & P             5, 6 Conjunction Addition
|   8.  S                 7, 2 Modus Ponens
|   9.  Q                    ACP, Case 2
|  10.  Q & R             9, 6 Conjunction Addition
|  11.  S                10, 3 Modus Ponens
12.     S                 8, 11  Exhaustive cases

John

My calculator said it, I believe it, that settles it

Answer by Edwin McCravy(20054) About Me  (Show Source):
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1. (PvQ)&R
2. (R&P)>S
3. (Q&R)>S
             Therefore, S
       | 4. ~S          Assumption for Indirect Proof
       | 5. R&(PvQ)     1, Commutation  
       | 6. R           5, Simplification
       | 7. ~(Q&R)      3,4 Modus Tollens
       | 8. ~Qv~R       7, DeMorgan's law  
       | 9. ~Rv~Q       8, Commutation
       |10. ~~R         6, Double negation
       |11. ~Q          9,10 Disjunctive syllogism 
       |12. PvQ         1, Simplification
       |13. QvP         12, Commutation  
       |14. P           13,11 Disjunctive syllogism
       |15. R&P         6,14 Conjunction
       |16. S           2,15 Modus ponens     
       |17. S&~S        16,4 Conjunction   <--(contradiction)
18. ~~S                 4-17 Indirect Proof
19. S                   18, Double negation 

Edwin

Answer by robertb(5830) About Me  (Show Source):
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1. (PvQ)&R ------------------Hypothesis
2. (R&P)>S ------------------Hypothesis
3. (Q&R)>S ------------------Hypothesis
4. (P&R)v(Q&R) --------------Distributivity on #1
5. (R&P)v(Q&R) --------------Commutativity on #4
6. S ----------------------Case Analysis based on #2, #3, and #5