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Proof step 268 from proof int2bytes-rec-pos-any (previous step)

n ≤ 8 ∧ 8 ≤ 8, # int2bits(j % 2 ^ (i * 8)) = (i→n(i) - 1) * 8 + n, 0 ≤ bytes2int(bits2bytes(firstn(n, int2bits(j % 2 ^ (i * 8)))) + bits2bytes(restn(n, int2bits(j % 2 ^ (i * 8))))), 8 < (i→n(i) - 1) * 8 + n, n < 8 + (i→n(i) - 1) * 8, n < (i→n(i) - 1) * 8 + 7, 2 ^ (i * 8) ≤ j, 0 ≤ j % 2 ^ (i * 8), n < 8, 0 < i ⊦ bits2bytes(firstn(8, mklist(false, 8 - n) + int2bits(j % 2 ^ (i * 8)))) + bits2bytes(restn((8 - 8) + n, int2bits(j % 2 ^ (i * 8)))) = bits2bytes(firstn(n, int2bits(j % 2 ^ (i * 8)))) + bits2bytes(restn(n, int2bits(j % 2 ^ (i * 8)))), n = 0, i→n(i) - 1 = 0, mklist(false, 8 - n) + int2bits(j % 2 ^ (i * 8)) = [], 1 + 8 * i→n(i) = (i→n(i) - 1) * 8 + n, (i→n(i) - 1) * 8 + n = 8 * i→n(i), i = 0, (i→n(i) - 1) * 8 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m * n = n * m;% ⊦ m < n → m ≤ n;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes1 + bytes0 ↔ bytes = bytes1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;% ⊦ i * j = j * i;