> To be pedantic, not guaranteed. The xor of multiple elements may erroneously have a passing checksum, resulting in an undetected false decode

The false decodes can be detected. During peeling, deleting a false decode inserts a new element with the opposite sign of count. Later, you decode this second false element and end up with the same element in both the A / B and B / A result sets (as long as decode completes without encountering a cycle).

So, after decode, check for any elements present in both A / B and B / A result sets and remove them.

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Beyond that, you can also use the cell position for additional checksum bits in the decode process without increasing the data structure's bit size. i.e., if we attempt to decode the element X from a cell at position m, then one of the h_i(x) hash functions for computing indices should return m.

There's even a paper about a variant of IBFs that has no checksum field at all: https://arxiv.org/abs/2211.03683. It uses the cell position among other techniques.