[smaudet]

I think I was misunderstood.

I do not mean H0(H1(m0)+H1(m1)) nor H1(m0)+H(m1) but Reinman(x={0,1000})(H0(m(x)))

Where there are 1000 hashes. So H0 must be broken one thousand times, then it does not matter that some attack exists to reduce the security 1000 times because the attack must be performed 1000 times. You could easily nest these so that F(y)=Reinman(x={0,y})(H0(m(y))) and take G(z)=Reinman(y={1,z})(F(y))

So that G(3) e.g. would produce 6 hashes of strength H0. No hash is taking another hash as a function, but the message itself here provides security - you must not just find a duplicate for one hash, but all 6 simultaneously. I wonder if the increased complexity might easily defeat most attacks on H0.