reserve Y for non empty set,
  a, b for Function of Y,BOOLEAN,
  G for Subset of PARTITIONS(Y),
  A, B for a_partition of Y;

theorem
  All(Ex(a,A,G),B,G) '<' Ex(Ex(a,B,G),A,G)
proof
A1: Ex(a,B,G) = B_SUP(a,CompF(B,G)) by BVFUNC_2:def 10;
  let z be Element of Y;
  assume
A2: (All(Ex(a,A,G),B,G)).z=TRUE;
A3: now
    assume not (for x being Element of Y st x in EqClass(z,CompF(B,G)) holds
    (Ex(a,A,G)).x=TRUE);
    then (B_INF(Ex(a,A,G),CompF(B,G))).z = FALSE by BVFUNC_1:def 16;
    hence contradiction by A2,BVFUNC_2:def 9;
  end;
A4: z in EqClass(z,CompF(B,G)) by EQREL_1:def 6;
  now assume
    not (ex x being Element of Y st x in EqClass(z,CompF(A,G)) & (a). x=TRUE);
    then (B_SUP(a,CompF(A,G))).z = FALSE by BVFUNC_1:def 17;
    then (Ex(a,A,G)).z=FALSE by BVFUNC_2:def 10;
    hence contradiction by A4,A3;
  end;
  then consider x1 being Element of Y such that
A5: x1 in EqClass(z,CompF(A,G)) and
A6: (a).x1=TRUE;
  x1 in EqClass(x1,CompF(B,G)) by EQREL_1:def 6;
  then (Ex(a,B,G)).x1=TRUE by A1,A6,BVFUNC_1:def 17;
  then (B_SUP(Ex(a,B,G),CompF(A,G))).z = TRUE by A5,BVFUNC_1:def 17;
  hence thesis by BVFUNC_2:def 10;
end;
