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Proof step 30 from proof int2bytes-len-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
0 ≤ i, 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1) ⊦ (1 + i→n(8 * i)) / 8 = i→n(i), (1 + i→n(8 * i)) % 8 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i + 0 = i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ ~ i < 0 ↔ 0 < i;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # mklist(boolvar, n) = n;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;%0 ≤ i ⊦ int2bits(~(2 ^ i)) = true ' + mklist(false, i→n(i));% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ 0 - i = ~ i;