How does name resolution work in compound "with" statements?

Question

Which instance of Ready gets tested in the following code, and why?

interface

type
  TObject1 = class
  ...
  public
    property Ready: boolean read FReady write FReady;
  end;

  TObject2 = class
  ...
  public
    property Ready: boolean read FReady write FReady;
  end;

implementation

var
  Object1: TObject1;
  Object2: TObject2;

...

procedure test;
var
  Ready: boolean;
begin
  Ready:= true;
  with Object1, Object2 do begin
    if Ready then ShowMessage('which one?');
  end; {with}
end;

Answer 1

The last one.

with Object1, Object2 do

is equivalent to

with Object1 do
  with Object2 do

and so Object2 will be the number-one priority.

The official documentation on this matter.