递归不可达序数

$(1){\psi }_I(0)=1st(1-{)}^{1,0}2$

$(2){\psi }_I(a+1)\ne {\psi }_I(a)$则${\psi }_I(a+1)=(1-{)}^{1,0}aft{\psi }_I(a)$