
theorem Th19:
  for A,B,C,x being set holds C c= A & not x in A implies ((id A)
  +*(B --> x))"(C \ {x}) = C \ B
proof
  let A,B,C,x be set;
  assume that
A1: C c= A and
A2: not x in A;
  not x in C \ B
  by A1,A2;
  then (C \ B) misses {x} by ZFMISC_1:50;
  then C \ B \ {x} = C \ B by XBOOLE_1:83;
  hence thesis by A1,Th16;
end;
