theorem Th6: phi in X implies phi is (X,{R#0(S)})-provable
proof
assume phi in X; then reconsider Xphi={phi} as Subset of X by ZFMISC_1:31;
{[{phi},phi]} is ({},{R#0(S)})-derivable; then
phi is (Xphi,{R#0(S)})-provable; hence thesis;
end;
