
theorem
for X be set, P be with_empty_element semi-diff-closed
      Subset-Family of X,
    M be pre-Measure of P,
    A,B be set st A in P & B in P & A \ B in P & B c= A holds
  M.A >= M.B
proof
   let X be set, P be with_empty_element semi-diff-closed
     Subset-Family of X,
       M be pre-Measure of P,
   A,B be set;
   assume that
A1: A in P & B in P & A \ B in P and
A2: B c= A;

   consider F be disjoint_valued FinSequence of P such that
A3:    A \ B = Union F by A1,SRINGS_3:def 1;

A7:rng <*B*> = {B} by FINSEQ_1:38; then
   reconsider G = <*B*> as disjoint_valued FinSequence of P
     by FINSEQ_1:def 4,A1,ZFMISC_1:31;

   now assume union rng G meets union rng F; then
    consider x be object such that
A4:  x in union rng G & x in union rng F by XBOOLE_0:3;
    consider P be set such that
A5:  x in P & P in rng G by A4,TARSKI:def 4;
    P in {B} by A5,FINSEQ_1:38; then
A6: x in B by A5,TARSKI:def 1;

    x in A \ B by A3,A4,CARD_3:def 4;
    hence contradiction by A6,XBOOLE_0:def 5;
   end; then
   reconsider H = G^F as disjoint_valued FinSequence of P by Th43;

A8:union rng G = B by A7,ZFMISC_1:25;

   rng H = rng G \/ rng F by FINSEQ_1:31; then
   union rng H = union rng G \/ union rng F by ZFMISC_1:78; then
   Union H = B \/ union rng F by A8,CARD_3:def 4
          .= B \/ (A \ B) by A3,CARD_3:def 4; then
   Union H = A \/ B by XBOOLE_1:39; then
   Union H = A by A2,XBOOLE_1:12; then
A9:M.A = Sum(M*H) by A1,Def8;

   Union G = B by A8,CARD_3:def 4; then
A10:M.B = Sum(M*G) by A1,Def8;

B0:now assume -infty in rng(M*G); then
    consider n be Element of NAT such that
B1:  n in dom(M*G) & -infty = (M*G).n by PARTFUN1:3;
    M.(G.n) = -infty by B1,FUNCT_1:12;
    hence contradiction by SUPINF_2:51;
   end;

A11:now assume -infty in rng(M*F); then
    consider n be Element of NAT such that
B2:  n in dom(M*F) & -infty = (M*F).n by PARTFUN1:3;
    M.(F.n) = -infty by B2,FUNCT_1:12;
    hence contradiction by SUPINF_2:51;
   end;
A12:now let n be Nat;
     assume n in dom(M*F); then
     (M*F).n = M.(F.n) & F.n in dom M by FUNCT_1:11,12;
     hence (M*F).n >= 0 by SUPINF_2:51;
    end;

   M*H = (M*G)^(M*F) by FINSEQOP:9; then
   Sum(M*H) = Sum(M*G) + Sum(M*F) by A11,B0,EXTREAL1:10;
   hence M.B <= M.A by A9,A10,A12,MESFUNC5:53,XXREAL_3:39;
end;
