YES

Problem:
 rev(a()) -> a()
 rev(b()) -> b()
 rev(++(x,y)) -> ++(rev(y),rev(x))
 rev(++(x,x)) -> rev(x)

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [++](x0, x1) = 2x0 + 2x1 + 7,
   
   [b] = 0,
   
   [rev](x0) = x0,
   
   [a] = 4
  orientation:
   rev(a()) = 4 >= 4 = a()
   
   rev(b()) = 0 >= 0 = b()
   
   rev(++(x,y)) = 2x + 2y + 7 >= 2x + 2y + 7 = ++(rev(y),rev(x))
   
   rev(++(x,x)) = 4x + 7 >= x = rev(x)
  problem:
   rev(a()) -> a()
   rev(b()) -> b()
   rev(++(x,y)) -> ++(rev(y),rev(x))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                   [1 0 0]          [0]
    [++](x0, x1) = [0 0 0]x0 + x1 + [0]
                   [0 0 1]          [1],
    
          [0]
    [b] = [1]
          [1],
    
                [1 0 1]     [0]
    [rev](x0) = [1 0 0]x0 + [1]
                [0 0 1]     [0],
    
          [0]
    [a] = [1]
          [1]
   orientation:
               [1]    [0]      
    rev(a()) = [1] >= [1] = a()
               [1]    [1]      
    
               [1]    [0]      
    rev(b()) = [1] >= [1] = b()
               [1]    [1]      
    
                   [1 0 1]    [1 0 1]    [1]    [1 0 1]    [1 0 1]    [0]                    
    rev(++(x,y)) = [1 0 0]x + [1 0 0]y + [1] >= [1 0 0]x + [0 0 0]y + [1] = ++(rev(y),rev(x))
                   [0 0 1]    [0 0 1]    [1]    [0 0 1]    [0 0 1]    [1]                    
   problem:
    
   Qed