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