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Proof step 22 from proof int2bytes-len-ls-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;