
theorem Th37:
  for S being non empty reflexive RelStr st for X being Subset of
S, x being Element of S holds x "/\" sup X = "\/"({x"/\"y where y is Element of
S: y in X},S) holds S is satisfying_MC
proof
  let S be non empty reflexive RelStr such that
A1: for X being Subset of S, x being Element of S holds x "/\" sup X =
  "\/"({x"/\"y where y is Element of S: y in X},S);
  let y be Element of S, D be non empty directed Subset of S;
  thus sup ({y} "/\" D) = "\/"({y"/\"z where z is Element of S: z in D},S) by
YELLOW_4:42
    .= y "/\" sup D by A1;
end;
