YES

Problem:
 \(x,x) -> e()
 /(x,x) -> e()
 .(e(),x) -> x
 .(x,e()) -> x
 \(e(),x) -> x
 /(x,e()) -> x
 .(x,\(x,y)) -> y
 .(/(y,x),x) -> y
 \(x,.(x,y)) -> y
 /(.(y,x),x) -> y
 /(x,\(y,x)) -> y
 \(/(x,y),x) -> y

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [.](x0, x1) = x0 + x1,
   
   [/](x0, x1) = x0 + 4x1 + 1,
   
   [e] = 1,
   
   [\](x0, x1) = 2x0 + 6x1 + 1
  orientation:
   \(x,x) = 8x + 1 >= 1 = e()
   
   /(x,x) = 5x + 1 >= 1 = e()
   
   .(e(),x) = x + 1 >= x = x
   
   .(x,e()) = x + 1 >= x = x
   
   \(e(),x) = 6x + 3 >= x = x
   
   /(x,e()) = x + 5 >= x = x
   
   .(x,\(x,y)) = 3x + 6y + 1 >= y = y
   
   .(/(y,x),x) = 5x + y + 1 >= y = y
   
   \(x,.(x,y)) = 8x + 6y + 1 >= y = y
   
   /(.(y,x),x) = 5x + y + 1 >= y = y
   
   /(x,\(y,x)) = 25x + 8y + 5 >= y = y
   
   \(/(x,y),x) = 8x + 8y + 3 >= y = y
  problem:
   \(x,x) -> e()
   /(x,x) -> e()
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                  [1 0 0]     [1 0 0]     [1]
    [/](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                  [0 0 0]     [0 0 0]     [0],
    
          [0]
    [e] = [0]
          [0],
    
                  [1 0 0]     [1 0 0]     [1]
    [\](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [0]
                  [0 0 0]     [0 0 0]     [0]
   orientation:
             [2 0 0]    [1]    [0]      
    \(x,x) = [0 0 0]x + [0] >= [0] = e()
             [0 0 0]    [0]    [0]      
    
             [2 0 0]    [1]    [0]      
    /(x,x) = [0 0 0]x + [0] >= [0] = e()
             [0 0 0]    [0]    [0]      
   problem:
    
   Qed