YES

Problem:
 strict:
  a() -> b()
 weak:
  f(s(x)) -> c(x,f(x))
  c(x,c(y(),z())) -> c(y(),c(x,z()))

Proof:
 Matrix Interpretation Processor: dim=5
  
  interpretation:
                 [1 0 0 0 0]     [1 0 0 0 1]     [0]
                 [1 0 0 0 0]     [0 1 0 0 1]     [0]
   [c](x0, x1) = [0 1 1 0 0]x0 + [0 0 1 0 0]x1 + [0]
                 [0 1 1 0 0]     [0 0 1 0 0]     [1]
                 [0 0 0 0 0]     [0 0 0 0 0]     [0],
   
         [0]
         [0]
   [b] = [0]
         [0]
         [0],
   
             [1 0 1 0 0]  
             [1 1 1 1 0]  
   [f](x0) = [0 0 0 1 0]x0
             [1 1 1 0 1]  
             [0 0 0 1 0]  ,
   
         [0]
         [1]
   [y] = [1]
         [0]
         [0],
   
         [1]
         [0]
   [a] = [0]
         [0]
         [1],
   
         [1]
         [1]
   [z] = [0]
         [0]
         [1],
   
             [1 0 1 0 1]     [0]
             [0 1 0 1 0]     [0]
   [s](x0) = [1 1 1 1 0]x0 + [0]
             [0 1 1 1 1]     [0]
             [0 0 0 0 0]     [1]
  orientation:
         [1]    [0]      
         [0]    [0]      
   a() = [0] >= [0] = b()
         [0]    [0]      
         [1]    [0]      
   
             [2 1 2 1 1]    [0]    [2 0 1 1 0]    [0]            
             [2 3 3 3 2]    [0]    [2 1 1 2 0]    [0]            
   f(s(x)) = [0 1 1 1 1]x + [0] >= [0 1 1 1 0]x + [0] = c(x,f(x))
             [2 2 2 2 1]    [1]    [0 1 1 1 0]    [1]            
             [0 1 1 1 1]    [0]    [0 0 0 0 0]    [0]            
   
                     [1 0 0 0 0]    [2]    [1 0 0 0 0]    [2]                  
                     [1 0 0 0 0]    [2]    [1 0 0 0 0]    [2]                  
   c(x,c(y(),z())) = [0 1 1 0 0]x + [2] >= [0 1 1 0 0]x + [2] = c(y(),c(x,z()))
                     [0 1 1 0 0]    [3]    [0 1 1 0 0]    [3]                  
                     [0 0 0 0 0]    [0]    [0 0 0 0 0]    [0]                  
  problem:
   strict:
    
   weak:
    f(s(x)) -> c(x,f(x))
    c(x,c(y(),z())) -> c(y(),c(x,z()))
  Qed