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;

theorem Th6:
for n,k be Element of NAT holds min* {n,k} = min(n,k) & min {n,k} = min(n,k)
proof
let n,k be Element of NAT;
A1: min {n} = n by Th5;
  {n,k}={n}\/{k} by ENUMSET1:1;
  then
A2: min {n,k} = min(min {n},min {k}) by Th2;
  min {n,k} = min* {n,k} by Th1;
  hence thesis by A2,A1,Th5;
end;
