theorem recneq3 (z: nat) (S: set) (_n1 _n2: nat): $ _n1 = _n2 -> recn z S _n1 = recn z S _n2 $;
_n1 = _n2 -> _n1 = _n2
_n1 = _n2 -> recn z S _n1 = recn z S _n2