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Proof step 42 from proof int2bytes-len-ls-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
# int2bits(i) / 8 = i→n(i0), # int2bits(i) ≤ i→n(8 * i0) + 1, 0 < i, i→n(8 * i0 - 1) + 1 < # int2bits(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: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ i * j = j * i;