theorem
  abs abs f = abs f
proof
  set f1 = abs f;
  thus
A1: dom abs f1 = dom abs f by Def36;
  let x be object;
  assume
A2: x in dom abs f1;
  hence (abs f1).x = abs(f1.x) by Def36
    .= abs(abs(f.x)) by A1,A2,Def36
    .= (abs f).x by A1,A2,Def36;
end;
