reserve X,Y for set, p,x,x1,x2,y,y1,y2,z,z1,z2 for object;
reserve f,g,h for Function;

theorem
  f c= g & f c= h implies g|dom f = h|dom f
proof
  assume that
A1: f c= g and
A2: f c= h;
  thus g|dom f = f by A1,Th21
    .= h|dom f by A2,Th21;
end;
