
theorem Th3:
  for A, B being set, f being Function, x being set holds A misses
  B implies (f +* (A --> x)).:B = f.:B
proof
  let A, B be set, f be Function, x be set;
  assume that
A1: A /\ B = {} and
A2: (f +* (A --> x)).:B <> f.:B;
A3: dom(f +* (A --> x)) = dom f \/ dom(A --> x) by FUNCT_4:def 1;
A4: dom(A --> x) = A by FUNCOP_1:13;
A5: not (for y being object holds y in (f +* (A --> x)).:B iff y in f.:B)
     by A2,TARSKI:2;
  now
    per cases by A5;
    case
      ex y being set st y in (f +* (A --> x)).:B & not y in f.:B;
      then consider y being set such that
A6:   y in (f +* (A --> x)).:B and
A7:   not y in f.:B;
      consider z being object such that
A8:   z in dom(f +* (A --> x)) and
A9:   z in B and
A10:  y = (f +* (A --> x)).z by A6,FUNCT_1:def 6;
      not z in A by A1,A9,XBOOLE_0:def 4;
      then z in dom f & y = f.z by A3,A4,A8,A10,FUNCT_4:11,XBOOLE_0:def 3;
      hence contradiction by A7,A9,FUNCT_1:def 6;
    end;
    case
      ex y being set st not y in (f +* (A --> x)).:B & y in f.:B;
      then consider y being set such that
A11:  not y in (f +* (A --> x)).:B and
A12:  y in f.:B;
      consider z being object such that
A13:  z in dom f and
A14:  z in B and
A15:  y = f.z by A12,FUNCT_1:def 6;
      not z in A by A1,A14,XBOOLE_0:def 4;
      then
A16:  y = (f +* (A --> x)).z by A4,A15,FUNCT_4:11;
      z in dom(f +* (A --> x)) by A3,A13,XBOOLE_0:def 3;
      hence contradiction by A11,A14,A16,FUNCT_1:def 6;
    end;
  end;
  hence thesis;
end;
