It seems this would only be a risk if the very last hashing function were compromised. Even then you would need to know how to generate the correct input to the last hashing function using the other hashing functions. It seems inherently more secure to me.
if hash1(x) has collisions, then so does hash2(hash1(x)) and hash4(hash3(... and so on, until we reach the full hashing stack, which also has collisions. Similarly, if hash2, hash3, etc, or hash11 have collisions, then so does X11(x). Simply put, if there's a collision attack for any hash#(x), then the same attack applies to X11(x).
This is the same baseless declaration with more words. You still failed to make the correlation or demonstrate a causation.
Why would this result in what is essentially a cascade failure? Why would an attack on Math A result in a failure of all other Math B to Math K to be Math anymore?
And on top of it, it's still an if...
I think you're fundamentally wrong. Collision on hash A does not break Hash B.
Really? This is very concerning that you seem to think this... None of what I am saying is baseless - for some reason you aren't even running simple tests?
Simple example:
sha256(crc32("plumless")) = sha256("4DDB0C25") = b7186aaec033aa3fb05e6b41444618f30e299f3879dde29d31f055f64fd8be49
sha256(crc32("buckeroo")) = sha256("4DDB0C25") = b7186aaec033aa3fb05e6b41444618f30e299f3879dde29d31f055f64fd8be49
crc32 has known collisions and is essentially broken in that sense, and so sha256(crc32()) is as well, and so blake256(kekkack(sha256(crc32(etc)) has known collisions.
Yes, this is an "if" situation... if a collision attack is discovered against any of the 11 sha3 candidate algorithms, that collision attack works for the whole chain. So, 11 chained algorithms are inherently less secure in regards to collision resistance vs 1 algorithm.
Demonstrate the part in bold. Everything else is proven and makes sense. Show us.