SOLUTION: Solve use reductio ad absurdum 1. ~P→(R∙S) 2. ~Q→(R∙T) 3. ~(S∨T) ∴ P∙Q

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Question 1117150: Solve use reductio ad absurdum
1. ~P→(R∙S)
2. ~Q→(R∙T)
3. ~(S∨T) ∴ P∙Q
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!


1. ~P —> (R*S)    Premise

2. ~Q —> (R*T)    Premise

3.  ~(SvT)        Premise    [ Use RA to show (P*Q) ]

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Ok, so we assume ~(P*Q) and show this leads to a contradiction (ANY contradiction). 

Once we get a contradiction, we can conclude ~~(P*Q) and then use Double Negation Elimination to get the ultimate conclusion P*Q.



Start a subproof for the Reductio ad Absurdum (RA) portion...

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4. |  ~(P*Q)             RA assumption

5. |  ~P v ~Q            4, DeMorgan's (DeM)

6. |  (R*S) v (R*T)      5,1,2 Constructive Dilemma (CD)

7. |  ~S * ~T            3 DeM

8. |  ~S                 7  AND eliminations (aka &E)

9. |   ~R v ~S           8  ADDition (aka vI)

10. |  ~(R*S)            9  DeM

11. |  ~T                7  &E

12. |  ~R v ~T           11  ADD  (vI)

13. |  ~(R*T)            12 DeM

14. | ~(R*S) * ~(R*T)    10,13 Conjunction

15. | ~[(R*S) v (R*T) ]  14 DeM  (contradicts line 6, ends subproof) 

16. ~~(P*Q)              4-15 RA  

17.  P*Q                 16  Double Negation Elimination (DNE)