YES

Problem:
 not(not(x)) -> x
 not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
 not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
                  [1 0]     [1 0]     [1]
   [or](x0, x1) = [0 3]x0 + [0 3]x1 + [1],
   
               [1 2]  
   [not](x0) = [0 1]x0,
   
                   [1 0]     [1 0]     [0]
   [and](x0, x1) = [0 3]x0 + [0 3]x1 + [1]
  orientation:
                 [1 4]          
   not(not(x)) = [0 1]x >= x = x
   
                  [1 6]    [1 6]    [3]    [1 6]    [1 6]    [0]                                         
   not(or(x,y)) = [0 3]x + [0 3]y + [1] >= [0 3]x + [0 3]y + [1] = and(not(not(not(x))),not(not(not(y))))
   
                   [1 6]    [1 6]    [2]    [1 6]    [1 6]    [1]                                        
   not(and(x,y)) = [0 3]x + [0 3]y + [1] >= [0 3]x + [0 3]y + [1] = or(not(not(not(x))),not(not(not(y))))
  problem:
   not(not(x)) -> x
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                [1 0 1]     [0]
    [not](x0) = [0 0 1]x0 + [0]
                [0 1 0]     [1]
   orientation:
                  [1 1 1]    [1]         
    not(not(x)) = [0 1 0]x + [1] >= x = x
                  [0 0 1]    [1]         
   problem:
    
   Qed