I suppose I missed the assumption that h_0, h_1 and h_2 cannot be trivially decomposed into concatenations of other hash functions, whereas h_3 and h_4 can. Al

I suppose I missed the assumption that h_0, h_1 and h_2 cannot be trivially decomposed into concatenations of other hash functions, whereas h_3 and h_4 can. Also, there's no obscurity since h_1 and h_2 are the actual targeted hash functions, h_3 and h_4 are just to show that concatenation doesn't change the actual problem or make it harder.