| > What would happen in practice? People would notice. A fork will be decided. There's nothing in the Bitcoin protocol that explicitly prevents a longer and valid chain from being accepted. Forks happen every day where miners are competing to construct the longest accepted chain on the network (it's part of the protocol). However, the difficulty in maliciously changing previous blocks in the Bitcoin blockchain is explicitly parameterized by the security level of the PoW mechanism. There are undergrad CS courses where this analysis is probably a homework problem. In PoS, immutability is baked into the protocol since the block selection algorithm doesn't explicitly cause forking as it does in PoW based blockchains. Decentralization creates trust that newly added blocks contain valid transactions, not to ensure that previous blocks don’t change. This is why the finality times of Bitcoin are strictly worse than the finality times of Algorand, or other PoS blockchains. Not because one network is more decentralized than the other, but because Bitcoin’s immutability property is hamstringed by the PoW security level. EDIT: Anticipating your response requesting citations - here are the lecture notes from the "Foundations of Blockchain Systems" course at the University of Washington, Seattle [1]. From section 3.2, I quote: "One property of the blockchain is immutability. This means that the chain cannot be changed given the last block in the chain. This is because it is difficult to find a hash of x that is equal to hash of y in computationally feasible time (collision-resistance). We add the qualifier “given the last block” because if we disagree on which is the last block, we may have problems. With distributed mining, there is a possibility of simultaneous mining, however these ties should be resolved quickly with proper tuning of parameters. Therefore, although we may not agree on the last block all of the time, all blocks will lead back to the genesis, starting block, and the large majority of the chain will be in agreement." [1] https://ece595uwseattle.github.io/Scribe_notes_Lecture_3.pdf |
> This is because it is difficult to find a hash of x that is equal to hash of y in computationally feasible time (collision-resistance).
This is exactly what I was explaining.