theorem
 for x,y being object holds
  rng(f+~(x,y)) c= rng f \ {x} \/ {y}
proof let x,y be object;
  per cases;
  suppose
A1: not x in rng f;
    then f+~(x,y) = f by Th103;
    then rng(f+~(x,y)) = rng f \ {x} by A1,ZFMISC_1:57;
    hence thesis by XBOOLE_1:7;
  end;
  suppose that
A2: x in rng f and
A3: x = y;
    f+~(x,y) = f by A3,Th102;
    hence thesis by A2,A3,ZFMISC_1:116;
  end;
  suppose that
A4: x <> y;
    not x in rng(f+~(x,y)) by A4,Th100;
    then
A5: rng(f+~(x,y)) \ {x} = rng(f+~(x,y)) by ZFMISC_1:57;
    rng(x .--> y) = {y} by FUNCOP_1:8;
    then
A6: rng f \/ rng((x .--> y)*f) c= rng f \/ {y} by RELAT_1:26,XBOOLE_1:9;
    rng(f+~(x,y)) c= rng f \/ rng((x .--> y)*f) by Th17;
    then rng(f+~(x,y)) c= rng f \/ {y} by A6;
    then rng(f+~(x,y)) c= rng f \/ {y} \ {x} by A5,XBOOLE_1:33;
    hence thesis by A4,ZFMISC_1:123;
  end;
end;
