YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [and](x0, x1) = 3x0 + x1 + 5,
   
   [or](x0, x1) = 3x0 + x1 + 3,
   
   [not](x0) = 4x0
  orientation:
   or(x,x) = 4x + 3 >= x = x
   
   and(x,x) = 4x + 5 >= x = x
   
   not(not(x)) = 16x >= x = x
   
   not(and(x,y)) = 12x + 4y + 20 >= 12x + 4y + 3 = or(not(x),not(y))
   
   not(or(x,y)) = 12x + 4y + 12 >= 12x + 4y + 5 = and(not(x),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