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Proof step 47 from proof int2bytes-len-ls-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 48:
# int2bits(i) / 8 = i→n(i0), i→n(8 * i0 - 1) < i→n(i0) * 8, n→i(n) ≤ i0, i < 2 ^ (n→i(n) * 8 - 1), 0 < i0, i→n(8 * i0 - 1) + 1 < # int2bits(i), 0 < i, # int2bits(i) ≤ 1 + i→n(i0) * 8 ⊦ # int2bits(i) % 8 = 0, i = 0, i0 = 0, n = 0

Rule argument

Additional information

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