simple relay question

I'm a newbie here, so excuse me if I overlooked any relevant posts that would answer this question.I have an '02 elantra with factory alarm/keyless. The factory alarm has no armed output and no LED. I want to add a Millenium FX pager, and a flashing LED. Replacing the factory system is not an option. Since there is no armed output, I want to create a "mock" armed output using the door locks. The elantra has a wire that tests (-) when the doors are unlocked, and dead when the doors are locked.I would like to tap into this lead to create a (-) output when:the doors are locked ANDthe ignition is offThis (-) output will be used to satisfy the pager's (-) armed input requirement and will also be used to drive a dummy flashing LED.Here's my proposed relay setup, which requires one relay to convert the dead signal to (-), and a second relay to inactivate the first relay when the ingnition is switched on. Am I doing something wrong? Is there a simpler way? I realize the first relay will be energized whenever the car is off and the doors is unlocked, but this is okay with me.

oops... i now see multiple mistakes. the armed (-) signal going to the pager transmitter should be connected to 87a, not 87 on the first relay, and the second relay is all wrong. i should have relay 2 inactivate the armed signal from 87 on relay 1 when switched 12V is applied.

well the problem lies in that the door locks only lock and unlock momentarily. so the relay would only come on and then off again. I am assuming that the door locks reverse polarity when they lock and when they unlock..? i wouldnt really advise using 2 relays to stay on while the vehicle is off as it would draw around .2 A... let me think about this one...

ok, let me see - if i'm reading you right, you have a wire that when the doors are unlocked, STAYS (-)  and what you are wanting is when the doors are locked to have something that stays (-)  --- in both cases i'm already assuming the ignition in off position ---  that is simple, and wont require any current being drawn just so long as the doors are locked.  once the doors are unlocked,  then the circuit will draw power (but we're assuming once you unlock your doors, you wont be long before you'll be driving your car, so that shouln't be a problem anyhow)...    your origional drawing is right on target, except like you said, the relay on the left, you need to move the wire from 87 to 87a.  the other relay is fine - it provides power (if necessary) to the relay on the left, in case a ground is applied to pin 86 while the ignition is off.  i'm assuming your pager alarm has an ACC wire on it that would disable it while the ignition is in the on position, so you don't need to worry about the position of either relay while driving.  if not, a 3rd relay could be added to further disable it.... but if you switch the right hand relay around, it will seem to work,  except that opening the door on your car wont trigger the alarm.