Question 1117150
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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)