Question 1189673
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1. ~S ∧ ~R                Premise  
2. ~S                     1, Simplification (SIMP)
3. ~R                     1, SIMP
4. ~R ∧ ~S                3,2 Conjunction Introduction (CI)
// Note, we could have just done Commutation of 1 to arrive at (~R ∧ ~S) in one step
5.:: S v R                Conditional Proof (CP) assumption #1
// Either S or R is true, let's assume S is true first...
6.:: S                    CP assumption #2 
7.:: ~(~R ∧ ~S)           Clearly (~R ∧ ~S) can not be true, because S is true.  The rule name here escapes me.
// Now assume R is true
8.:: R                    CP assumption #3
9.:: ~(~R ∧ ~S)           Clearly (~R ∧ ~S) can not be true, because R is true.  Same rule as line 7.
10.:: (S v R) --> ~(~R ∧ ~S)     5-9, CP
11.:: (~R ∧ ~S) --> ~(S v R)     10, Transposition (if A --> B then ~B --> ~A)
12. (~R ∧ ~S) --> ~(S v R)       5-11, CP
*DONE*