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 Th25:
  for B1,B2 being set st B1 in Sigma & B2 in Sigma holds for C1,C2
  being thin of P holds B1 \/ C1 = B2 \/ C2 implies P.B1 = P.B2
proof
  let B1,B2 be set;
  assume
A1: B1 in Sigma & B2 in Sigma;
  let C1,C2 be thin of P;
  assume
A2: B1 \/ C1 = B2 \/ C2;
  then
A3: B1 c= B2 \/ C2 by XBOOLE_1:7;
A4: B2 c= B1 \/ C1 by A2,XBOOLE_1:7;
  consider D1 being set such that
A5: D1 in Sigma and
A6: C1 c= D1 and
A7: P.D1 = 0 by Def4;
A8: B1 \/ C1 c= B1 \/ D1 by A6,XBOOLE_1:9;
  consider D2 being set such that
A9: D2 in Sigma and
A10: C2 c= D2 and
A11: P.D2 = 0 by Def4;
A12: B2 \/ C2 c= B2 \/ D2 by A10,XBOOLE_1:9;
  reconsider B1,B2,D1,D2 as Event of Sigma by A1,A5,A9;
A13: P.(B1 \/ D1) <= P.B1 + P.D1 by PROB_1:39;
  P.B2 <= P.(B1 \/ D1) by A4,A8,PROB_1:34,XBOOLE_1:1;
  then
A14: P.B2 <= P.B1 by A7,A13,XXREAL_0:2;
A15: P.(B2 \/ D2) <= P.B2 + P.D2 by PROB_1:39;
  P.B1 <= P.(B2 \/ D2) by A3,A12,PROB_1:34,XBOOLE_1:1;
  then P.B1 <= P.B2 by A11,A15,XXREAL_0:2;
  hence thesis by A14,XXREAL_0:1;
end;
