
theorem Th1:
  for A, B be non empty set, A1, A2 being non empty Subset of A st
    A1 misses A2
  for f1 being Function of A1,B, f2 being Function of A2,B holds
  f1|(A1 /\ A2) = f2|(A1 /\ A2) &
  (f1 union f2)|A1 = f1 & (f1 union f2)|A2 = f2
proof
  let A, B be non empty set, A1, A2 be non empty Subset of A;
  assume A1:A1 misses A2;
  let f1 be Function of A1,B, f2 be Function of A2,B;
  A1 /\ A2 c= A2 by XBOOLE_1:17;
  then reconsider g2 = f2|(A1 /\ A2) as Function of {},B by A1,FUNCT_2:32;
  A1 /\ A2 c= A1 by XBOOLE_1:17;
  then reconsider g1 = f1|(A1 /\ A2) as Function of {},B by A1,FUNCT_2:32;
  g1 = g2;
  hence thesis by Def1;
end;
