YES

Problem:
 f(0()) -> cons(0())
 f(s(0())) -> f(p(s(0())))
 p(s(X)) -> X

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]  
   [p](x0) = [1 1 0]x0
             [0 0 1]  ,
   
             [1 0 1]     [0]
   [f](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 1 0]     [0]
   [s](x0) = [0 0 0]x0 + [1]
             [0 0 1]     [0],
   
         [0]
   [0] = [0]
         [1],
   
                [1 0 0]     [0]
   [cons](x0) = [0 0 0]x0 + [1]
                [0 0 0]     [0]
  orientation:
            [1]    [0]            
   f(0()) = [1] >= [1] = cons(0())
            [0]    [0]            
   
               [1]    [1]               
   f(s(0())) = [1] >= [1] = f(p(s(0())))
               [0]    [0]               
   
             [1 1 0]    [0]         
   p(s(X)) = [1 1 0]X + [1] >= X = X
             [0 0 1]    [0]         
  problem:
   f(s(0())) -> f(p(s(0())))
   p(s(X)) -> X
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]  
    [p](x0) = [0 0 1]x0
              [0 1 0]  ,
    
              [1 0 1]  
    [f](x0) = [0 0 1]x0
              [0 0 0]  ,
    
              [1 0 0]     [0]
    [s](x0) = [0 0 1]x0 + [0]
              [0 1 0]     [1],
    
          [0]
    [0] = [0]
          [0]
   orientation:
                [1]    [0]               
    f(s(0())) = [1] >= [0] = f(p(s(0())))
                [0]    [0]               
    
                  [0]         
    p(s(X)) = X + [1] >= X = X
                  [0]         
   problem:
    p(s(X)) -> X
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                    [1]
     [p](x0) = x0 + [0]
                    [0],
     
                    [0]
     [s](x0) = x0 + [1]
                    [0]
    orientation:
                   [1]         
     p(s(X)) = X + [1] >= X = X
                   [0]         
    problem:
     
    Qed