theorem Th6:
  'not' ( p 'or' q ) => ( 'not' p '&' 'not' q ) in TAUT(A)
proof
  'not' ( p 'or' q ) = 'not' 'not' ( 'not' p '&' 'not' q ) by QC_LANG2:def 3;
  hence thesis by LUKASI_1:25;
end;
