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Proof step 72 from proof int2bytes-rec-neg-05 (previous step)

0 ≤ i * 8 - 1 ∧ ~(2 ^ ((i * 8 - 1) + 1)) ≤ j ∧ j < ~(2 ^ (i * 8 - 1)), i→n((i * 8 - 1) + 2) % 8 = 0, j < 0, 0 < 2 ^ (i * 8) + j, 0 < j + 2 ^ (i * 8), j + 2 ^ (i * 8 - 1) < 0, 0 < i ⊦ bits2bytes(int2bits(j)) = i2b(~1) ' + fix(i→n(i) + 1, bits2bytes(int2bits(j))).rest, j = 0, i→n((i * 8 - 1) + 2) / 8 = i→n(i) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
i→n(1 + i * 8) % 8 = 0, 0 ≤ 2 ^ (i * 8) + j, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ i→n(1 + i * 8) / 8 = i→n(i) + 1, j = 0, bits2bytes(int2bits(j)) = i2b(~1) ' + fix(i→n(i) + 1, bits2bytes(int2bits(j))).rest

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ 2 ^ (i - 1) < 2 ^ i ↔ 0 ≤ i;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i < ~ j ↔ i + j < 0;% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ i + (j - i) = j;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i * j = j * i;