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:
 DP Processor:
  DPs:
   .#(.(x,y),z) -> .#(y,z)
   .#(.(x,y),z) -> .#(x,.(y,z))
   i#(.(x,y)) -> i#(x)
   i#(.(x,y)) -> i#(y)
   i#(.(x,y)) -> .#(i(y),i(x))
  TRS:
   .(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))
  Matrix Interpretation Processor: dim=1
   
   usable rules:
    .(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))
   interpretation:
    [i#](x0) = 4x0,
    
    [.#](x0, x1) = x0,
    
    [i](x0) = 4x0 + 5,
    
    [.](x0, x1) = x0 + x1 + 3,
    
    [1] = 6
   orientation:
    .#(.(x,y),z) = x + y + 3 >= y = .#(y,z)
    
    .#(.(x,y),z) = x + y + 3 >= x = .#(x,.(y,z))
    
    i#(.(x,y)) = 4x + 4y + 12 >= 4x = i#(x)
    
    i#(.(x,y)) = 4x + 4y + 12 >= 4y = i#(y)
    
    i#(.(x,y)) = 4x + 4y + 12 >= 4y + 5 = .#(i(y),i(x))
    
    .(1(),x) = x + 9 >= x = x
    
    .(x,1()) = x + 9 >= x = x
    
    .(i(x),x) = 5x + 8 >= 6 = 1()
    
    .(x,i(x)) = 5x + 8 >= 6 = 1()
    
    i(1()) = 29 >= 6 = 1()
    
    i(i(x)) = 16x + 25 >= 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 + 6 >= x + y + z + 6 = .(x,.(y,z))
    
    i(.(x,y)) = 4x + 4y + 17 >= 4x + 4y + 13 = .(i(y),i(x))
   problem:
    DPs:
     
    TRS:
     .(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))
   Qed