YES

Problem:
 implies(not(x),y) -> or(x,y)
 implies(not(x),or(y,z)) -> implies(y,or(x,z))
 implies(x,or(y,z)) -> or(y,implies(x,z))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [implies](x0, x1) = 5x0 + 5x1 + 1,
   
   [not](x0) = 5x0 + 1,
   
   [or](x0, x1) = 5x0 + x1
  orientation:
   implies(not(x),y) = 25x + 5y + 6 >= 5x + y = or(x,y)
   
   implies(not(x),or(y,z)) = 25x + 25y + 5z + 6 >= 25x + 5y + 5z + 1 = implies(y,or(x,z))
   
   implies(x,or(y,z)) = 5x + 25y + 5z + 1 >= 5x + 5y + 5z + 1 = or(y,implies(x,z))
  problem:
   implies(x,or(y,z)) -> or(y,implies(x,z))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                        [1 0 0]     [1 1 1]     [0]
    [implies](x0, x1) = [1 0 0]x0 + [0 0 1]x1 + [1]
                        [1 0 0]     [0 1 0]     [0],
    
                   [1 0 0]     [1 0 0]     [0]
    [or](x0, x1) = [0 0 0]x0 + [0 0 1]x1 + [1]
                   [0 0 0]     [0 1 0]     [0]
   orientation:
                         [1 0 0]    [1 0 0]    [1 1 1]    [1]    [1 0 0]    [1 0 0]    [1 1 1]    [0]                     
    implies(x,or(y,z)) = [1 0 0]x + [0 0 0]y + [0 1 0]z + [1] >= [1 0 0]x + [0 0 0]y + [0 1 0]z + [1] = or(y,implies(x,z))
                         [1 0 0]    [0 0 0]    [0 0 1]    [1]    [1 0 0]    [0 0 0]    [0 0 1]    [1]                     
   problem:
    
   Qed