reserve Y for non empty set;
reserve Y for non empty set;
reserve Y for non empty set;

theorem
  for a,b being Function of Y,BOOLEAN holds 'not'( a 'eqv' b) = (a
  'eqv' 'not' b)
proof
  let a,b be Function of Y,BOOLEAN;
    let x be Element of Y;
    ('not'( a 'eqv' b)).x =('not'( (a 'imp' b) '&' (b 'imp' a))).x by
BVFUNC_4:7
      .=('not'( ('not' a 'or' b) '&' (b 'imp' a))).x by BVFUNC_4:8
      .=('not'( ('not' a 'or' b) '&' ('not' b 'or' a))).x by BVFUNC_4:8
      .=('not'('not' a 'or' b) 'or' 'not'('not' b 'or' a)).x by BVFUNC_1:14
      .=(('not' 'not' a '&' 'not' b) 'or' 'not'('not' b 'or' a)).x by
BVFUNC_1:13
      .=((a '&' 'not' b) 'or' ('not' 'not' b '&' 'not' a)).x by BVFUNC_1:13
      .=(((a '&' 'not' b) 'or' b) '&' ((a '&' 'not' b) 'or' 'not' a)).x by
BVFUNC_1:11
      .=(((a 'or' b) '&' ('not' b 'or' b)) '&' ((a '&' 'not' b) 'or' 'not' a
    )).x by BVFUNC_1:11
      .=(((a 'or' b) '&' ('not' b 'or' b)) '&' ((a 'or' 'not' a) '&' ('not'
    b 'or' 'not' a))).x by BVFUNC_1:11
      .=(((a 'or' b) '&' I_el(Y)) '&' ((a 'or' 'not' a) '&' ('not' b 'or'
    'not' a))).x by BVFUNC_4:6
      .=(((a 'or' b) '&' I_el(Y)) '&' (I_el(Y) '&' ('not' b 'or' 'not' a))).
    x by BVFUNC_4:6
      .=((a 'or' b) '&' (I_el(Y) '&' ('not' b 'or' 'not' a))).x by BVFUNC_1:6
      .=(('not' a 'or' 'not' b) '&' ('not' 'not' b 'or' a)).x by BVFUNC_1:6
      .=(('not' a 'or' 'not' b) '&' ('not' b 'imp' a)).x by BVFUNC_4:8
      .=((a 'imp' 'not' b) '&' ('not' b 'imp' a)).x by BVFUNC_4:8
      .=(a 'eqv' 'not' b).x by BVFUNC_4:7;
    hence thesis;
end;
