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*From*: Stephan Merz <stephan.merz@xxxxxxxxx>*Date*: Sun, 15 Dec 2019 14:38:51 +0100*References*: <2b6263b2-50cc-4591-b748-ea66d0b23bd8@googlegroups.com>

Hello, a trivial error in your definition, and probably the one that TLC tells you about, is that you pass the set containing the only element 1..Len(seq) – which itself is a set of integers – to the Random operator, which will therefore return that set, and the evaluation of the _expression_ `j<i' in the definition of Remove fails because < expects two integers. You clearly meant to write Remove(Random(1..Len(seq)), seq) without the set brackets around the argument to Random. However, the more fundamental issue is that you cannot define a function whose outcome is random: a function is deterministic and always returns the same result for the same input. In particular, the CHOOSE operator yields an arbitrary but fixed element: no matter how often you repeat the experiment, TLC will always compute the same sequence. If this were not so, TLA+ would violate the elementary reflexivity law \A x : x = x of first-order logic. Also, I'm assuming that you do not really mean "random" in the sense of probability theory, but that you intend to write a specification that allows any execution in which an arbitrary element of your original sequence is removed. We just saw that this cannot be done using just functions and TLA+ operators, but you can instead use a relational _expression_ and for example write something like Next == \/ ... \/ \E i \in 1 .. Len(seq) : seq' = Remove(i, seq) where your specification has a variable seq that represents the current system state. Hope this helps, Stephan
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**References**:**[tlaplus] Eliminate random element from seq***From:*p . tollot

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