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Proof step 93 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) < # int2bits(j % 2 ^ (i * 8)) / 8 + 1, 2 ^ (i * 8) ≤ j, 0 < i, i→n(i) + 1 < # int2bits(j % 2 ^ (i * 8)) / 8 + 1 ⊦ 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 = # int2bits(j % 2 ^ (i * 8)) / 8 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 94:
# int2bits(j % 2 ^ (i * 8)) = i→n(i * 8) + 1, # int2bits(j % 2 ^ (i * 8)).rest % 8 = 0, 0 < i, 2 ^ (i * 8) ≤ j, i→n(i) < (i→n(i * 8) + 1) / 8 ⊦ i→n(i) = (i→n(i * 8) + 1) / 8, bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = lastn(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest, int2bits(j % 2 ^ (i * 8)).rest = [], (i→n(i * 8) + 1) % 8 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ m ≠ n → (m < n + 1 ↔ m < n);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;