YES

Problem:
 strict:
  p(0(),y) -> y
  p(s(x),y) -> s(p(x,y))
 weak:
  p(x,y) -> p(x,s(y))

Proof:
 Matrix Interpretation Processor: dim=5
  
  interpretation:
                 [1 0 0 0 0]     [1 0 0 0 0]  
                 [1 0 0 0 0]     [1 1 1 1 1]  
   [p](x0, x1) = [1 0 0 0 0]x0 + [0 0 1 1 0]x1
                 [0 0 0 0 0]     [0 1 0 1 0]  
                 [0 0 0 0 0]     [0 1 0 1 1]  ,
   
         [1]
         [0]
   [0] = [0]
         [0]
         [0],
   
             [1 0 0 0 0]  
             [0 1 0 0 0]  
   [s](x0) = [0 0 1 0 0]x0
             [0 0 0 0 0]  
             [0 0 0 1 0]  
  orientation:
              [1 0 0 0 0]    [1]         
              [1 1 1 1 1]    [1]         
   p(0(),y) = [0 0 1 1 0]y + [1] >= y = y
              [0 1 0 1 0]    [0]         
              [0 1 0 1 1]    [0]         
   
               [1 0 0 0 0]    [1 0 0 0 0]     [1 0 0 0 0]    [1 0 0 0 0]             
               [1 0 0 0 0]    [1 1 1 1 1]     [1 0 0 0 0]    [1 1 1 1 1]             
   p(s(x),y) = [1 0 0 0 0]x + [0 0 1 1 0]y >= [1 0 0 0 0]x + [0 0 1 1 0]y = s(p(x,y))
               [0 0 0 0 0]    [0 1 0 1 0]     [0 0 0 0 0]    [0 0 0 0 0]             
               [0 0 0 0 0]    [0 1 0 1 1]     [0 0 0 0 0]    [0 1 0 1 0]             
   
            [1 0 0 0 0]    [1 0 0 0 0]     [1 0 0 0 0]    [1 0 0 0 0]             
            [1 0 0 0 0]    [1 1 1 1 1]     [1 0 0 0 0]    [1 1 1 1 0]             
   p(x,y) = [1 0 0 0 0]x + [0 0 1 1 0]y >= [1 0 0 0 0]x + [0 0 1 0 0]y = p(x,s(y))
            [0 0 0 0 0]    [0 1 0 1 0]     [0 0 0 0 0]    [0 1 0 0 0]             
            [0 0 0 0 0]    [0 1 0 1 1]     [0 0 0 0 0]    [0 1 0 1 0]             
  problem:
   strict:
    p(s(x),y) -> s(p(x,y))
   weak:
    p(x,y) -> p(x,s(y))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
                  [1 1]     [2 0]  
    [p](x0, x1) = [0 1]x0 + [1 0]x1,
    
                   [0]
    [s](x0) = x0 + [1]
   orientation:
                [1 1]    [2 0]    [1]    [1 1]    [2 0]    [0]            
    p(s(x),y) = [0 1]x + [1 0]y + [1] >= [0 1]x + [1 0]y + [1] = s(p(x,y))
    
             [1 1]    [2 0]     [1 1]    [2 0]             
    p(x,y) = [0 1]x + [1 0]y >= [0 1]x + [1 0]y = p(x,s(y))
   problem:
    strict:
     
    weak:
     p(x,y) -> p(x,s(y))
   Qed