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Proof step 33 from proof int2bytes-rec-neg-05 (previous step)

(8 * i→n(i) + 8) % 8 = 0, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0, 8 * i→n(i) + 8 ≤ 8 ⊦ j = 0, i2b(j) = i2b(~1), j = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
8 * i→n(i) = 0, 0 < i, j + 2 ^ (i * 8 - 1) < 0, 0 < j + 2 ^ (i * 8), 0 < 2 ^ (i * 8) + j, j < 0 ⊦ j = 0, i2b(j) = i2b(~1), j = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ m + n ≤ m ↔ n = 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ 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;