Are lenient public source licenses eligible?

I’m specifically referring of the FlowCrypt Android app: GitHub - FlowCrypt/flowcrypt-android: FlowCrypt Android App

The source code is public and modification and redistribution are permitted by the license. However, there are some restrictions in place that forbid certain kinds of modification and redistribution, so the license is not pure FOSS.

Here is the license itself: https://github.com/FlowCrypt/flowcrypt-android/blob/master/LICENSE

Would it be okay with F-Droid if FlowCrypt were to be published on F-Droid?

Thanks!

IANAL, that is a proprietary license.

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According to the inclusion policy, all apps in F-Droid must be licensed under an FSF-approved free software license.

The FlowCrypt license is demonstrably not a free software license, as it does not grant freedom 0, the freedom to use the program for any purpose. The license places several restrictions on usage:

Use Limitation: This software is free to use with the following limitations:
 - Size of any sent email message including attachments cannot exceed 5MB.
 - Password encrypted messages and attachments have to expire in 3 days.
 - Messages, attachments or files can only be transferred through used email provider or through services provided by Licensor. Never through 3rd party or custom services.
 - Plain-text portion of password encrypted messages must only include a link to a page on flowcrypt.com domain. Recipient cannot be directed to any other place or means of decrypting password encrypted messages.
 - Password encrypted messages cannot link or lead to a web page where encrypted reply is possible.
 - Cannot request attestation of secondary send-from email addresses (email aliases).
 - Limit one re-attestation (change of associated pubkey) per email address.
 - Cannot set up a custom email footer to be included in sent messages (encrypted or not).
 - Maximum 100 email addresses per domain (except gmail.com, outlook.com and other public email service provider domains excluded).
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Okay, that answers that, then. Thanks for the response!

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