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Proof step 130 from proof int2bytes-rec-pos-any (previous step)

# int2bits(j % 2 ^ (i * 8)) % 8 = 0 ∧ int2bits(j % 2 ^ (i * 8)) ≠ [], i→n(i * 8) % 8 = 0, # int2bits(j % 2 ^ (i * 8)) = i→n(i * 8), bytes2int(bits2bytes(int2bits(j % 2 ^ (i * 8)))) < 0, # int2bits(j % 2 ^ (i * 8)) / 8 < i→n(i), 0 ≤ j % 2 ^ (i * 8), 0 < i, 2 ^ (i * 8) ≤ j ⊦ i→n(i) = # int2bits(j % 2 ^ (i * 8)) / 8, i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 131:
# int2bits(j % 2 ^ (i * 8)) = i→n(i * 8), i→n(i * 8) % 8 = 0, 2 ^ (i * 8) ≤ j, 0 < i, 0 ≤ j % 2 ^ (i * 8), i→n(i * 8) / 8 < i→n(i), bytes2int(bits2bytes(int2bits(j % 2 ^ (i * 8)))) < 0 ⊦ i = 0, i→n(i) = i→n(i * 8) / 8

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ i * j = j * i;