theorem sublistAteq1 (_n1 _n2 L1 L2: nat): $ _n1 = _n2 -> (sublistAt _n1 L1 L2 <-> sublistAt _n2 L1 L2) $;
_n1 = _n2 -> _n1 = _n2
_n1 = _n2 -> (sublistAt _n1 L1 L2 <-> sublistAt _n2 L1 L2)