reserve n,m,k for Element of NAT,
  x,X for set,
  A1 for SetSequence of X,
  Si for SigmaField of X,
  XSeq for SetSequence of Si;
reserve Omega for non empty set,
  Sigma for SigmaField of Omega,
  ASeq for SetSequence of Sigma,
  P for Probability of Sigma;

theorem Th16:
  A1 is non-descending implies (Partial_Diff_Union A1).0 = A1.0 &
  for n being Nat holds (Partial_Diff_Union A1).(n+1) = A1.(n+1) \ A1.n
proof
  assume
A1: A1 is non-descending;
  thus (Partial_Diff_Union A1).0 = A1.0 by PROB_3:def 3;
    let n be Nat;
    thus (Partial_Diff_Union A1).(n+1) = A1.(n+1) \ (Partial_Union A1).n by
PROB_3:def 3
      .= A1.(n+1) \ A1.n by A1,Lm2;
end;
