reserve Y for non empty set;
reserve Y for non empty set;
reserve Y for non empty set;
reserve Y for non empty set,
  a,b,c,d,e,f,g for Function of Y,BOOLEAN;

theorem
  (a 'imp' b) '&' 'not' b '<' 'not' a
proof
  let z be Element of Y;
  reconsider bz = b.z as boolean object;
A1: ((a 'imp' b) '&' 'not' b).z =(a 'imp' b).z '&' ('not' b).z by
MARGREL1:def 20
    .=('not' a 'or' b).z '&' ('not' b).z by BVFUNC_4:8
    .=('not' b).z '&' (('not' a).z 'or' b.z) by BVFUNC_1:def 4
    .=('not' b).z '&' ('not' a).z 'or' ('not' b).z '&' b.z by XBOOLEAN:8
    .=('not' b).z '&' ('not' a).z 'or' 'not' bz '&' bz by MARGREL1:def 19
    .=('not' b).z '&' ('not' a).z 'or' FALSE by XBOOLEAN:138
    .=('not' b).z '&' ('not' a).z;
  assume
A2: ((a 'imp' b) '&' 'not' b).z=TRUE;
  now
    assume ('not' a).z<>TRUE;
    then ('not' b).z '&' ('not' a).z =FALSE '&' ('not' b).z by XBOOLEAN:def 3
      .=FALSE;
    hence contradiction by A2,A1;
  end;
  hence thesis;
end;
