https://gcc.gnu.org/bugzilla/show_bug.cgi?id=93131
--- Comment #6 from Andrew Pinski <pinskia at gcc dot gnu.org> --- Note the generic version of this is: ((A & N) == CST1) & ((A & M) == CST2) if (N&M)&CST1 == (N&M)&CST2, then (A&(N|M)) == (CST1|CST2) else false