That is true ONLY if you need to open both devices (mosfets) at the same time BUT it is NOT necessary for simple switch. For simple switch you have to open only one device and you'll get current flow through opened device! That means if you remove LEDs from your schematic, with high level on control signal one of MOSFETS MUST BE opened. Which one depends on the current polarity on switch terminals. And that is what he wants.As I mentioned earlier, when using two devices on opposite polarities you need two control signals of opposite polarities, as each device needs it's own base or gate supply relative to it's own emitter or source.
What is the range of voltage of your controlling signal? If you need final and optimal solution I suggest to tell us what you really want. Examples with LEDs are not good. Why?
If your controlling signal can open MOSFET then it probably can provide 20mA which is enough to turn LED on. So connect leds as they are connected on schematic with ne555 and in the middle connect your controlling signal and that's it.
If you need latching such state (active state "stays" while controlling signal has changed) you must use flip-flop.
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