YES

Problem:
 p(s(x)) -> x
 fac(0()) -> s(0())
 fac(s(x)) -> times(s(x),fac(p(s(x))))

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
                     [1 0]       
   [times](x0, x1) = [0 0]x0 + x1,
   
               [2 2]     [0]
   [fac](x0) = [1 0]x0 + [3],
   
         [3]
   [0] = [0],
   
             [1 0]  
   [p](x0) = [1 0]x0,
   
             [1 1]  
   [s](x0) = [2 2]x0
  orientation:
             [1 1]          
   p(s(x)) = [1 1]x >= x = x
   
              [6]    [3]         
   fac(0()) = [6] >= [6] = s(0())
   
               [6 6]    [0]    [5 5]    [0]                           
   fac(s(x)) = [1 1]x + [3] >= [1 1]x + [3] = times(s(x),fac(p(s(x))))
  problem:
   p(s(x)) -> x
   fac(s(x)) -> times(s(x),fac(p(s(x))))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
                      [1 0]     [2 0]  
    [times](x0, x1) = [2 0]x0 + [0 0]x1,
    
                [1 2]     [1]
    [fac](x0) = [2 0]x0 + [0],
    
              [1 0]     [1]
    [p](x0) = [1 0]x0 + [0],
    
              [1 1]     [0]
    [s](x0) = [3 3]x0 + [3]
   orientation:
              [1 1]    [1]         
    p(s(x)) = [1 1]x + [0] >= x = x
    
                [7 7]    [7]    [7 7]    [4]                           
    fac(s(x)) = [2 2]x + [0] >= [2 2]x + [0] = times(s(x),fac(p(s(x))))
   problem:
    
   Qed