theorem Th62:
  a 'or' (b 'nor' c) = (a 'or' 'not' b) '&' (a 'or' 'not' c)
proof
  thus a 'or' (b 'nor' c) =a 'or' 'not' (b 'or' c) by Th2
    .=a 'or' ('not' b '&' 'not' c) by BVFUNC_1:13
    .=(a 'or' 'not' b) '&' (a 'or' 'not' c) by BVFUNC_1:11;
end;
