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Proof step 90 from proof int2bytes-rec-pos-any (previous step)

¬ # int2bits(j % 2 ^ (i * 8)) % 8 ≠ 0, # int2bits(j % 2 ^ (i * 8)).rest % 8 = 0, # int2bits(j % 2 ^ (i * 8)) = i→n(i * 8) + 1, i→n(i) < # bits2bytes(int2bits(j % 2 ^ (i * 8))), 2 ^ (i * 8) ≤ j, 0 < i, i→n(i) + 1 < # bits2bytes(int2bits(j % 2 ^ (i * 8))) ⊦ int2bits(j % 2 ^ (i * 8)).rest = [], bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = lastn(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest, i→n(i) + 1 = # bits2bytes(int2bits(j % 2 ^ (i * 8)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 91:
(i→n(i * 8) + 1) % 8 = 0, # int2bits(j % 2 ^ (i * 8)).rest % 8 = 0, # int2bits(j % 2 ^ (i * 8)) = i→n(i * 8) + 1, i→n(i) < # bits2bytes(int2bits(j % 2 ^ (i * 8))), 2 ^ (i * 8) ≤ j, 0 < i, i→n(i) + 1 < # bits2bytes(int2bits(j % 2 ^ (i * 8))) ⊦ int2bits(j % 2 ^ (i * 8)).rest = [], bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = lastn(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest, i→n(i) + 1 = # bits2bytes(int2bits(j % 2 ^ (i * 8)))

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ i * j = j * i;