theorem Th64:
  a 'eqv' (b 'nor' c) = (a 'or' b 'or' c) '&' ('not' a 'or' 'not' (b 'or' c))
proof
  thus a 'eqv' (b 'nor' c) =a 'eqv' 'not' (b 'or' c) by Th2
    .=(a 'or' 'not' 'not' (b 'or' c)) '&' ('not' a 'or' 'not' (b 'or' c))
      by BVFUNC_6:91
    .=(a 'or' b 'or' c) '&' ('not' a 'or' 'not' (b 'or' c)) by BVFUNC_1:8;
end;
