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