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*From*: Victor Miraldo <victor.miraldo@xxxxxxxx>*Date*: Fri, 13 Aug 2021 06:12:03 -0700 (PDT)*Ironport-hdrordr*: A9a23:brg6mKN0g0WWT8BcTl6jsMiBIKoaSvp037BL7TEKdfUxSKb0qynApoV+6faZskd3ZJnP8erwTJVpbxvnhOtICMoqTMaftDCPghrbEGga1/qU/9SCIVyEygbpvp0QDZSWdueAdGSS1vyKnzVQeuxIqLL3kpxA492us0uFYjsEV0gK1XYANu/0KDwReOGGbaBJd6Z0JfAom9NjQxgqhnTRPAh3YwEOnbz2fGKMW296O/fv0mn+6A+A2frBChCdmj0eXzlMzbpn0W+AvRf++rzLiYDK9iPh

Hello all,

I have some TLA+ spec that makes heavy use of functions and TLC takes a long time (minutes) to check that even the simplest property, i.e., `[](FALSE)`, doesn't hold. This made me question how TLC handles functions, but I can't find information elsewhere.

When I write [ x \in TypeX |-> f(x) ], is TLC generating all elements in `TypeX` then computing their image under `f` or does it wait until its called with an argument `m` to compute its image under `f`?

Is there a way to make TLA+ lazy, preventing it from ever generating `TypeX` in its entirety? Unfortunately we have some very large `TypeX`s in our generated code, such as the type of functions that receive lists and return lists.

Unfortunately, I can't use operators here because these need to be passed as arguments to mutually recursive operators.

Thanks,

Victor

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**Follow-Ups**:**Re: [tlaplus] How does TLC functions work under the hood?***From:*Markus Kuppe

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