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Proof step 280 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(i2b(bits2int(firstn(n, int2bits(j % 2 ^ (i * 8))))) ' + bits2bytes(restn(n, int2bits(j % 2 ^ (i * 8))))), 0 < i, n < 8, 0 ≤ j % 2 ^ (i * 8), 2 ^ (i * 8) ≤ j, n < (i→n(i) - 1) * 8 + 7, n < 8 + (i→n(i) - 1) * 8, 8 < (i→n(i) - 1) * 8 + n ⊦ bits2int(mklist(false, 8 - n) + firstn((8 - 8) + n, int2bits(j % 2 ^ (i * 8)))) = bits2int(firstn(n, int2bits(j % 2 ^ (i * 8)))), i2b(bits2int(mklist(false, 8 - n) + firstn((8 - 8) + n, int2bits(j % 2 ^ (i * 8))))) = i2b(bits2int(firstn(n, int2bits(j % 2 ^ (i * 8))))), (i→n(i) - 1) * 8 = 1, i = 0, (i→n(i) - 1) * 8 + n = 8 * i→n(i), 1 + 8 * i→n(i) = (i→n(i) - 1) * 8 + n, mklist(false, 8 - n) + int2bits(j % 2 ^ (i * 8)) = [], i→n(i) - 1 = 0, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i2b(i) ≠ i2b(j) → i ≠ j; (forward)% ⊦ m * n = n * m;% ⊦ m < n → m ≤ n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;% ⊦ i * j = j * i;