YES

Problem:
 strict:
  f(x,c(y)) -> f(x,s(f(y,y)))
  f(s(x),y) -> f(x,s(c(y)))
 weak:
  rand(x) -> x
  rand(x) -> rand(s(x))

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
                 [2 0]     [1 1]     [1]
   [f](x0, x1) = [0 0]x0 + [0 0]x1 + [2],
   
                [2 0]     [2]
   [rand](x0) = [0 3]x0 + [0],
   
             [1 1]     [0]
   [c](x0) = [2 1]x0 + [2],
   
             [1 0]  
   [s](x0) = [0 0]x0
  orientation:
               [2 0]    [3 2]    [3]    [2 0]    [3 1]    [2]                 
   f(x,c(y)) = [0 0]x + [0 0]y + [2] >= [0 0]x + [0 0]y + [2] = f(x,s(f(y,y)))
   
               [2 0]    [1 1]    [1]    [2 0]    [1 1]    [1]               
   f(s(x),y) = [0 0]x + [0 0]y + [2] >= [0 0]x + [0 0]y + [2] = f(x,s(c(y)))
   
             [2 0]    [2]         
   rand(x) = [0 3]x + [0] >= x = x
   
             [2 0]    [2]    [2 0]    [2]             
   rand(x) = [0 3]x + [0] >= [0 0]x + [0] = rand(s(x))
  problem:
   strict:
    f(s(x),y) -> f(x,s(c(y)))
   weak:
    rand(x) -> rand(s(x))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
                  [1 1]     [1 0]  
    [f](x0, x1) = [1 1]x0 + [1 0]x1,
    
                 [1 0]     [0]
    [rand](x0) = [2 0]x0 + [2],
    
              [1 0]     [1]
    [c](x0) = [0 0]x0 + [0],
    
                   [0]
    [s](x0) = x0 + [2]
   orientation:
                [1 1]    [1 0]    [2]    [1 1]    [1 0]    [1]               
    f(s(x),y) = [1 1]x + [1 0]y + [2] >= [1 1]x + [1 0]y + [1] = f(x,s(c(y)))
    
              [1 0]    [0]    [1 0]    [0]             
    rand(x) = [2 0]x + [2] >= [2 0]x + [2] = rand(s(x))
   problem:
    strict:
     
    weak:
     rand(x) -> rand(s(x))
   Qed