reserve Y for non empty set,
  G for Subset of PARTITIONS(Y),
  a,b,c,u for Function of Y,BOOLEAN,
  PA for a_partition of Y;

theorem
  All(a,PA,G) '<' Ex(b,PA,G) 'imp' Ex(a '&' b,PA,G)
proof
  let z be Element of Y;
  assume
A1: All(a,PA,G).z=TRUE;
A2: now
    assume not (for x being Element of Y st x in EqClass(z,CompF(PA,G)) holds
    a.x=TRUE);
    then B_INF(a,CompF(PA,G)).z = FALSE by BVFUNC_1:def 16;
    hence contradiction by A1,BVFUNC_2:def 9;
  end;
  per cases;
  suppose
A3: Ex(b,PA,G).z=TRUE;
    now
      assume
      not (ex x being Element of Y st x in EqClass(z,CompF(PA,G)) & b. x=TRUE);
      then B_SUP(b,CompF(PA,G)).z = FALSE by BVFUNC_1:def 17;
      hence contradiction by A3,BVFUNC_2:def 10;
    end;
    then consider x1 being Element of Y such that
A4: x1 in EqClass(z,CompF(PA,G)) and
A5: b.x1=TRUE;
    (a '&' b).x1 =a.x1 '&' b.x1 by MARGREL1:def 20
      .=TRUE '&' TRUE by A2,A4,A5
      .=TRUE;
    then B_SUP(a '&' b,CompF(PA,G)).z = TRUE by A4,BVFUNC_1:def 17;
    then Ex(a '&' b,PA,G).z=TRUE by BVFUNC_2:def 10;
    hence
    (Ex(b,PA,G) 'imp' Ex(a '&' b,PA,G)).z =('not' Ex(b,PA,G).z) 'or' TRUE
    by BVFUNC_1:def 8
      .=TRUE by BINARITH:10;
  end;
  suppose
    Ex(b,PA,G).z<>TRUE;
    then Ex(b,PA,G).z=FALSE by XBOOLEAN:def 3;
    hence
    (Ex(b,PA,G) 'imp' Ex(a '&' b,PA,G)).z =('not' FALSE) 'or' Ex(a '&' b,
    PA,G).z by BVFUNC_1:def 8
      .=TRUE 'or' Ex(a '&' b,PA,G).z by MARGREL1:11
      .=TRUE by BINARITH:10;
  end;
end;
