YES

Problem:
 rec[even_0][1](0()) -> True()
 rec[even_0][1](S(0())) -> False()
 rec[even_0][1](S(S(x2))) -> rec[even_0][1](x2)
 main(x1) -> rec[even_0][1](x1)

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
                [1 0 1]  
   [main](x0) = [0 0 1]x0
                [1 0 0]  ,
   
             [0]
   [False] = [0]
             [0],
   
             [1 0 0]     [0]
   [S](x0) = [0 0 1]x0 + [0]
             [0 1 0]     [1],
   
            [0]
   [True] = [0]
            [0],
   
                          [1 0 1]  
   [rec[even_0][1]](x0) = [0 0 0]x0
                          [0 0 0]  ,
   
         [0]
   [0] = [0]
         [1]
  orientation:
                         [1]    [0]         
   rec[even_0][1](0()) = [0] >= [0] = True()
                         [0]    [0]         
   
                            [1]    [0]          
   rec[even_0][1](S(0())) = [0] >= [0] = False()
                            [0]    [0]          
   
                              [1 0 1]     [1]    [1 0 1]                       
   rec[even_0][1](S(S(x2))) = [0 0 0]x2 + [0] >= [0 0 0]x2 = rec[even_0][1](x2)
                              [0 0 0]     [0]    [0 0 0]                       
   
              [1 0 1]      [1 0 1]                       
   main(x1) = [0 0 1]x1 >= [0 0 0]x1 = rec[even_0][1](x1)
              [1 0 0]      [0 0 0]                       
  problem:
   main(x1) -> rec[even_0][1](x1)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                 [1 0 0]     [1]
    [main](x0) = [0 0 0]x0 + [0]
                 [0 0 1]     [0],
    
                           [1 0 0]  
    [rec[even_0][1]](x0) = [0 0 0]x0
                           [0 0 0]  
   orientation:
               [1 0 0]     [1]    [1 0 0]                       
    main(x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 = rec[even_0][1](x1)
               [0 0 1]     [0]    [0 0 0]                       
   problem:
    
   Qed