theorem eqmeq3 (n a _b1 _b2: nat): $ _b1 = _b2 -> (mod(n): a = _b1 <-> mod(n): a = _b2) $;
_b1 = _b2 -> _b1 = _b2
_b1 = _b2 -> (mod(n): a = _b1 <-> mod(n): a = _b2)