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From the yellow paper; for a block to be accepted as valid, an 8 byte number, n_rand, must be found that satisfies equation 253:

= block header without n_rand and MixHash
= difficulty
= nonce

My interpretation is that d is the dataset - a value cryptographically derived from the number of previous blocks. From this, the Mixhash is calculated in the PoW function

Why are the dataset and MixHash values required?

If the network were to agree to accept the PoW function with d set to 0, wouldn't the system still be cryptographically dependent on valid state transistions, as there must be consensus on H_n? Would the system not, therefore, still work?

From the yellow paper; for a block to be accepted as valid, an 8 byte number, n_rand, must be found that satisfies equation 253:

= block header without n_rand and MixHash
= difficulty
= nonce

My interpretation is that d is the dataset - a value cryptographically derived from the number of previous blocks. From this, the Mixhash is calculated in the PoW function

Why are the dataset and MixHash values required?

If the network were to agree to accept the PoW function with d set to 0, wouldn't the system still be cryptographically dependent on valid state transistions, as there must be consensus on H_n? Would the system not, therefore, still work?

From the yellow paper; for a block to be accepted as valid, an 8 byte number, n_rand, must be found that satisfies equation 253:

PoW(H_n,n_rand,d)[1] _< 2^256/H_d

H_n = block header without n_rand and MixHash
H_d = difficulty
n_rand = nonce

My interpretation is that d is the dataset - a value cryptographically derived from the number of previous blocks. From this, the Mixhash is calculated in the PoW function

Why are the dataset and MixHash values required?

If the network were to agree to accept the PoW function with d set to 0, wouldn't the system still be cryptographically dependent on valid state transistions, as there must be consensus on H_n? Would the system not, therefore, still work?

From the yellow paper; for a block to be accepted as valid, an 8 byte number, n_rand, must be found that satisfies equation 253:

= block header without n_rand and MixHash
= difficulty
= nonce

My interpretation is that d is the dataset - a value cryptographically derived from the number of previous blocks. From this, the Mixhash is calculated in the PoW function

Why are the dataset and MixHash values required?

If the network were to agree to accept the PoW function with d set to 0, wouldn't the system still be cryptographically dependent on valid state transistions, as there must be consensus on H_n? Would the system not, therefore, still work?