reserve k, l, m, n, i, j for Nat,
  K, N for non empty Subset of NAT,
  Ke, Ne, Me for Subset of NAT,
  X,Y for set;
reserve f for Function of Segm n,Segm k;
reserve x,y for set;

theorem Th50:
  3! = 6 & 4! = 24
proof
  thus
A1: 3! = (2+1)! .= 2*3 by NEWTON:14,15
    .= 6;
  thus 4! = (3+1)! .= 6*4 by A1,NEWTON:15
    .= 24;
end;
