YES

Problem:
 rec[flip_0][1](E()) -> E()
 rec[flip_0][1](Z(x1)) -> O(rec[flip_0][1](x1))
 rec[flip_0][1](O(x1)) -> Z(rec[flip_0][1](x1))
 main(x1) -> rec[flip_0][1](x1)

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [main](x0) = x0 + 1,
   
   [O](x0) = x0,
   
   [Z](x0) = x0,
   
   [rec[flip_0][1]](x0) = x0,
   
   [E] = 0
  orientation:
   rec[flip_0][1](E()) = 0 >= 0 = E()
   
   rec[flip_0][1](Z(x1)) = x1 >= x1 = O(rec[flip_0][1](x1))
   
   rec[flip_0][1](O(x1)) = x1 >= x1 = Z(rec[flip_0][1](x1))
   
   main(x1) = x1 + 1 >= x1 = rec[flip_0][1](x1)
  problem:
   rec[flip_0][1](E()) -> E()
   rec[flip_0][1](Z(x1)) -> O(rec[flip_0][1](x1))
   rec[flip_0][1](O(x1)) -> Z(rec[flip_0][1](x1))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]     [0]
    [O](x0) = [0 0 0]x0 + [0]
              [0 1 1]     [1],
    
              [1 0 0]     [0]
    [Z](x0) = [0 0 0]x0 + [0]
              [0 1 1]     [1],
    
                           [1 1 1]  
    [rec[flip_0][1]](x0) = [0 1 0]x0
                           [1 0 1]  ,
    
          [0]
    [E] = [1]
          [0]
   orientation:
                          [1]    [0]      
    rec[flip_0][1](E()) = [1] >= [1] = E()
                          [0]    [0]      
    
                            [1 1 1]     [1]    [1 1 1]     [0]                        
    rec[flip_0][1](Z(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = O(rec[flip_0][1](x1))
                            [1 1 1]     [1]    [1 1 1]     [1]                        
    
                            [1 1 1]     [1]    [1 1 1]     [0]                        
    rec[flip_0][1](O(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = Z(rec[flip_0][1](x1))
                            [1 1 1]     [1]    [1 1 1]     [1]                        
   problem:
    
   Qed