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Proof step 44 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 * i→n(i) + 8, 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, 8 < i→n((i * 8 - 1) + 2) ⊦ j = 0, i = 0, i2b(bits2int(firstn(8, int2bits(j)))) = i2b(~1), bits2int(firstn(8, int2bits(j))) = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
i→n(1 + i * 8) = 8 * i→n(i) + 8, 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 ⊦ bits2int(firstn(8, int2bits(j))) = ~1, i2b(bits2int(firstn(8, int2bits(j)))) = i2b(~1), i = 0, j = 0

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;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;%n ≠ 0 ⊦ (m + n0 * n) % n = m % n;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;%0 ≤ i ⊦ (* m) < i→n(i) ↔ n→i(m) < i;% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 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;