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' a 'imp' c) '<' b 'or' c
proof
  let z be Element of Y;
A1: ((a 'imp' b) '&' ('not' a 'imp' c)).z =(('not' a 'or' b) '&' ('not' a
  'imp' c)).z by BVFUNC_4:8
    .=(('not' a 'or' b) '&' ('not' 'not' a 'or' c)).z by BVFUNC_4:8
    .=('not' a 'or' b).z '&' (a 'or' c).z by MARGREL1:def 20
    .=(('not' a).z 'or' b.z) '&' (a 'or' c).z by BVFUNC_1:def 4
    .=(('not' a).z 'or' b.z) '&' (a.z 'or' c.z) by BVFUNC_1:def 4;
  assume
A2: ((a 'imp' b) '&' ('not' a 'imp' c)).z=TRUE;
  now
    reconsider az = a.z as boolean object;
    assume (b 'or' c).z<>TRUE;
    then (b 'or' c).z=FALSE by XBOOLEAN:def 3;
    then
A3: b.z 'or' c.z=FALSE by BVFUNC_1:def 4;
    c.z=TRUE or c.z=FALSE by XBOOLEAN:def 3;
    then (('not' a).z 'or' b.z) '&' (a.z 'or' c.z) ='not' az '&' az by A3,
MARGREL1:def 19
      .=FALSE by XBOOLEAN:138;
    hence contradiction by A2,A1;
  end;
  hence thesis;
end;
