theorem
  for p,q being Function, A being set holds dom p c= A & dom q misses A
  implies (p +* q)|A = p
proof
  let p,q be Function, A be set;
  assume that
A1: dom p c= A and
A2: dom q misses A;
  thus (p +* q )|A = p|A by A2,Th72
    .= p by A1,RELAT_1:68;
end;
