
theorem Th14:
  for A,B,x being set holds dom((id A)+*(B --> x)) = A \/ B
proof
  let A,B,x be set;
  dom((id A)+*(B --> x)) = dom (id A) \/ dom (B --> x) by FUNCT_4:def 1;
  then dom((id A)+*(B --> x)) = A \/ dom (B --> x);
  hence thesis by FUNCOP_1:13;
end;
