There are no requirements in ADS to disconnect for a L-N fault.But splutter splutter, why isn't a Zs generally necessary? Not saying you're wrong but curious to know your reasoning as to why you are not really bothered about ensuring the MCB L-N fault is within ADS requirements?
ADS is section 411, in chapter 41: protection against electric shock. L to N faults don't generally cause electric shock, as they cause no rise in voltage on exposed conductive parts, so the disconnection times in 411 do not apply. There are other reasons to disconnect (adiabatic), but these faults aren't required to meet the times tabled in 41.1
411.3.2.1
Except as provided by Regulation 411.3.2.5, a protective device shall automatically interrupt the supply to the line conductor of a circuit or equipment in the event of a fault of negligible impedance between the line conductor and an exposed-conductive-part or a protective conductor in the circuit or equipment within the disconnection time required by Regulation 411.3.2.2, 411.3.2.3 or 411.3.2.4.
(* L to N faults don't generally cause electric shock, as they cause no rise in voltage on exposed conductive parts: actually, I believe that this isn't the full story. I believe that a L-N fault would see a rise in voltage on the earthing system for a TN-C-S earthing arrangement, the E and N being connected at the cutout.
However, a fault so low that it failed to disconnect the protective device quickly would be a long way downstream of the cutout. I believe that much of the voltage would have dropped between the fault and the cutout, so the rise on the earthing system would be low.)