  Thus Val(L(ω1,ω1)) is not definable even by an L(ω1,ω1)-formula, so a fortiori L(ω1,ω1) is incomplete. Similar arguments show that Scott's undefinability theorem continues to hold when ω1 is replaced by any successor cardinal κ+; accordingly the languages  L(κ+,κ+) are all incomplete.[7]