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Proof step 44 from proof int2bytes-len-neg (previous step)

# int2bits(j) = m * 8 + n ∧ n < 8 ∧ 8 ≠ 0, i→n(i) * 8 < i→n(8 * (i + 1) - 1), j < ~1, 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1), 0 ≤ i, # int2bits(j) ≤ i→n(8 * (i + 1) - 1) + 1, j < 0, 1 + i→n(i) * 8 < # int2bits(j) ⊦ j + 2 ^ (8 * i) = 0, m = i→n(i), n = 0, 8 * (i + 1) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
# int2bits(j) = m * 8 + n, 1 + i→n(i) * 8 < m * 8 + n, j < 0, m * 8 + n ≤ i→n(8 * (i + 1) - 1) + 1, 0 ≤ i, 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1), j < ~1, i→n(i) * 8 < i→n(8 * (i + 1) - 1), n < 8 ⊦ 8 * (i + 1) = 1, n = 0, m = i→n(i), j + 2 ^ (8 * i) = 0

Rule argument

Additional information

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