YES

Problem:
 .(1(),x) -> x
 .(x,1()) -> x
 .(i(x),x) -> 1()
 .(x,i(x)) -> 1()
 i(1()) -> 1()
 i(i(x)) -> x
 .(i(y),.(y,z)) -> z
 .(y,.(i(y),z)) -> z
 .(.(x,y),z) -> .(x,.(y,z))
 i(.(x,y)) -> .(i(y),i(x))

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [i](x0) = 4x0 + 1,
   
   [.](x0, x1) = x0 + x1 + 5,
   
   [1] = 6
  orientation:
   .(1(),x) = x + 11 >= x = x
   
   .(x,1()) = x + 11 >= x = x
   
   .(i(x),x) = 5x + 6 >= 6 = 1()
   
   .(x,i(x)) = 5x + 6 >= 6 = 1()
   
   i(1()) = 25 >= 6 = 1()
   
   i(i(x)) = 16x + 5 >= x = x
   
   .(i(y),.(y,z)) = 5y + z + 11 >= z = z
   
   .(y,.(i(y),z)) = 5y + z + 11 >= z = z
   
   .(.(x,y),z) = x + y + z + 10 >= x + y + z + 10 = .(x,.(y,z))
   
   i(.(x,y)) = 4x + 4y + 21 >= 4x + 4y + 7 = .(i(y),i(x))
  problem:
   .(i(x),x) -> 1()
   .(x,i(x)) -> 1()
   .(.(x,y),z) -> .(x,.(y,z))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [i](x0) = 4x0,
    
    [.](x0, x1) = 4x0 + x1 + 2,
    
    [1] = 2
   orientation:
    .(i(x),x) = 17x + 2 >= 2 = 1()
    
    .(x,i(x)) = 8x + 2 >= 2 = 1()
    
    .(.(x,y),z) = 16x + 4y + z + 10 >= 4x + 4y + z + 4 = .(x,.(y,z))
   problem:
    .(i(x),x) -> 1()
    .(x,i(x)) -> 1()
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 0]     [1]
     [i](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [0],
     
                   [1 0 0]     [1 0 0]  
     [.](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 0 0]     [0 0 0]  ,
     
           [0]
     [1] = [0]
           [0]
    orientation:
                 [2 0 0]    [1]    [0]      
     .(i(x),x) = [0 0 0]x + [0] >= [0] = 1()
                 [0 0 0]    [0]    [0]      
     
                 [2 0 0]    [1]    [0]      
     .(x,i(x)) = [0 0 0]x + [0] >= [0] = 1()
                 [0 0 0]    [0]    [0]      
    problem:
     
    Qed