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Proof step 33 from proof int2bytes-len-ls-01 (previous step)

0 ≤ i ∧ 1 < 1 + 8 * i→n(i0), # int2bits(i) / 8 = i→n(i0), 2 ^ (n→i(1 + 8 * i→n(i0)) - 2) ≤ i ∧ i < 2 ^ (n→i(1 + 8 * i→n(i0)) - 1), 0 < i, 0 ≤ i0 ⊦ # int2bits(i) % 8 = 0, i = 0, n ≤ i→n(i0) ↔ ¬ i < 2 ^ (n→i(n) * 8 - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
# int2bits(i) / 8 = i→n(i0), i < 2 ^ (8 * i0), 2 ^ (8 * i0 - 1) ≤ i, 0 < i, 0 < i0 ⊦ i0 = 0, n ≤ i→n(i0) ↔ ¬ i < 2 ^ (n→i(n) * 8 - 1), i = 0, # int2bits(i) % 8 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 2 ^ (i - 1) < 2 ^ i ↔ 0 ≤ i;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ 1 < 2 ^ i ↔ 0 < i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i + 0 = i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ 0 ≤ i → n→i(i→n(i)) = i;% ⊦ n→i(m * n) = n→i(m) * n→i(n);% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i * j = j * i;