theorem
  a 'nand' (a 'eqv' b) = a 'imp' (a 'xor' b)
proof
  a 'nand' (a 'eqv' b) = 'not' (a '&' (a 'eqv' b)) by th1
    .= 'not' a 'or' 'not' (a 'eqv' b) by BVFUNC_1:14
    .= 'not' a 'or' 'not' 'not' (a 'xor' b) by Th12
    .= 'not' a 'or' (a 'xor' b);
  hence thesis by BVFUNC_4:8;
end;
