Full RBF

[LoyceV]

And in such case, most of the stolen coin goes to miner as TX fee.

That's the beauty of it: it makes hunting compromised private keys futile

[satscraper]

The transaction B must pay 1000 satoshi more for its own bandwidth and since the transaction A paid 500 satoshi, it should pay at least 1500 satoshi in total.
So, as mentioend above by BlackHatCoiner, the fee rate for the replacement transaction must be at least 1500/1000 = 1.5 sat/vbyte.

Wow, completely forgot about A.   Nevertheless that is not a rule punctually kept. Miners may  accept and include into block transaction with the feerate literally equal to zero. They do it sometimes when this involves either  their own transactions (RBF included) or payment receiving (in cash for instance) outside the blockchain.

[DaveF]

And in such case, most of the stolen coin goes to miner as TX fee.

That's the beauty of it: it makes hunting compromised private keys futile

Yes, but people will still do it.

Full RBF will be enabled by default (for all users who haven't explicitly turned it off) in the next Bitcoin Core release, v28.0.

https://github.com/bitcoin/bitcoin/pull/30493

The other thing I still find interesting is looking at https://bitnodes.io/nodes/ and seeing how many old versions are still running out there.
We are talking about 28.0 now and yet more then 1/3 of the reachable nodes are on 25 or earlier.

What we don't know from that is how many are nodes that are used by people for transactions and how many are just sitting out there.

-Dave