YES

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

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
                [1 0]     [2]
   [rand](x0) = [0 2]x0 + [0],
   
                      [1 0]     [0]
   [c](x0, x1) = x0 + [0 0]x1 + [2],
   
             [2 1]  
   [g](x0) = [0 1]x0,
   
               
   [s](x0) = x0,
   
             [2 1]  
   [f](x0) = [0 1]x0
  orientation:
                  [2 1]    [2 0]    [2]    [2 1]    [2 0]    [2]               
   f(c(s(x),y)) = [0 1]x + [0 0]y + [2] >= [0 1]x + [0 0]y + [2] = f(c(x,s(y)))
   
                     [2 1]    [2 0]    [2]    [2 1]    [2 0]    [2]            
   f(c(s(x),s(y))) = [0 1]x + [0 0]y + [2] >= [0 1]x + [0 0]y + [2] = g(c(x,y))
   
                  [2 1]    [2 0]    [2]    [2 1]    [2 0]    [2]               
   g(c(x,s(y))) = [0 1]x + [0 0]y + [2] >= [0 1]x + [0 0]y + [2] = g(c(s(x),y))
   
                     [2 1]    [2 0]    [2]    [2 1]    [2 0]    [2]            
   g(c(s(x),s(y))) = [0 1]x + [0 0]y + [2] >= [0 1]x + [0 0]y + [2] = f(c(x,y))
   
             [1 0]    [2]         
   rand(x) = [0 2]x + [0] >= x = x
   
             [1 0]    [2]    [1 0]    [2]             
   rand(x) = [0 2]x + [0] >= [0 2]x + [0] = rand(s(x))
  problem:
   strict:
    f(c(s(x),y)) -> f(c(x,s(y)))
    f(c(s(x),s(y))) -> g(c(x,y))
    g(c(x,s(y))) -> g(c(s(x),y))
    g(c(s(x),s(y))) -> f(c(x,y))
   weak:
    rand(x) -> rand(s(x))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
                 [2 0]     [0]
    [rand](x0) = [0 0]x0 + [2],
    
                  [2 1]     [2 1]     [1]
    [c](x0, x1) = [0 0]x0 + [2 0]x1 + [0],
    
              [1 0]  
    [g](x0) = [0 0]x0,
    
                   [0]
    [s](x0) = x0 + [1],
    
              [1 0]  
    [f](x0) = [0 0]x0
   orientation:
                   [2 1]    [2 1]    [2]    [2 1]    [2 1]    [2]               
    f(c(s(x),y)) = [0 0]x + [0 0]y + [0] >= [0 0]x + [0 0]y + [0] = f(c(x,s(y)))
    
                      [2 1]    [2 1]    [3]    [2 1]    [2 1]    [1]            
    f(c(s(x),s(y))) = [0 0]x + [0 0]y + [0] >= [0 0]x + [0 0]y + [0] = g(c(x,y))
    
                   [2 1]    [2 1]    [2]    [2 1]    [2 1]    [2]               
    g(c(x,s(y))) = [0 0]x + [0 0]y + [0] >= [0 0]x + [0 0]y + [0] = g(c(s(x),y))
    
                      [2 1]    [2 1]    [3]    [2 1]    [2 1]    [1]            
    g(c(s(x),s(y))) = [0 0]x + [0 0]y + [0] >= [0 0]x + [0 0]y + [0] = f(c(x,y))
    
              [2 0]    [0]    [2 0]    [0]             
    rand(x) = [0 0]x + [2] >= [0 0]x + [2] = rand(s(x))
   problem:
    strict:
     f(c(s(x),y)) -> f(c(x,s(y)))
     g(c(x,s(y))) -> g(c(s(x),y))
    weak:
     rand(x) -> rand(s(x))
   Bounds Processor:
    bound: 1
    enrichment: match-rt
    automaton:
     final states: {6}
     transitions:
      c0(6,6) -> 6*
      rand1(9) -> 6*
      g0(6) -> 6*
      s0(6) -> 6*
      s1(9) -> 9*
      s1(6) -> 9*
      f0(6) -> 6*
      rand0(6) -> 6*
      c1(6,9) -> 10*
      f1(10) -> 6*
    problem:
     strict:
      g(c(x,s(y))) -> g(c(s(x),y))
     weak:
      rand(x) -> rand(s(x))
    Bounds Processor:
     bound: 1
     enrichment: match-rt
     automaton:
      final states: {5}
      transitions:
       c0(5,5) -> 5*
       rand1(10) -> 11*
       g0(5) -> 5*
       s0(5) -> 5*
       g1(9) -> 5*
       s1(12) -> 13*
       s1(5) -> 8*
       s1(8) -> 8*
       rand0(5) -> 5*
       c1(8,5) -> 9*
       11 -> 5*
       13 -> 10*
       8 -> 10*
       10 -> 12*
     problem:
      strict:
       
      weak:
       rand(x) -> rand(s(x))
     Qed