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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [.](x0, x1) = x0 + x1 + 3,
   
   [1] = 6,
   
   [i](x0) = 2x0 + 3
  orientation:
   .(1(),x) = x + 9 >= x = x
   
   .(x,1()) = x + 9 >= x = x
   
   .(i(x),x) = 3x + 6 >= 6 = 1()
   
   .(x,i(x)) = 3x + 6 >= 6 = 1()
   
   i(1()) = 15 >= 6 = 1()
   
   i(i(x)) = 4x + 9 >= x = x
   
   .(i(y),.(y,z)) = 3y + z + 9 >= z = z
   
   .(y,.(i(y),z)) = 3y + z + 9 >= z = z
  problem:
   .(i(x),x) -> 1()
   .(x,i(x)) -> 1()
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [.](x0, x1) = 6x0 + 4x1,
    
    [1] = 0,
    
    [i](x0) = 4x0 + 3
   orientation:
    .(i(x),x) = 28x + 18 >= 0 = 1()
    
    .(x,i(x)) = 22x + 12 >= 0 = 1()
   problem:
    
   Qed