theorem Th6:
  [#]A \ p is multiplicatively-closed
  proof
    reconsider p as prime Ideal of A by Lm5;
    reconsider M = [#]A \ p as Subset of A;
A1: not 1.A in p by IDEAL_1:19; then
    reconsider M as non empty Subset of A by XBOOLE_0:def 5;
    for a,b be Element of A st a in M & b in M holds a * b in M
    proof
      let a,b be Element of A;
      assume
A2:   a in M & b in M;
      assume not a*b in M; then
      a*b in p by XBOOLE_0:def 5; then
      a in p or b in p by RING_1:def 1;
      hence contradiction by A2,XBOOLE_0:def 5;
    end;
    hence thesis by A1,XBOOLE_0:def 5;
  end;
