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