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Proof step 176 from proof int2bytes-rec-neg-05 (previous step)

# int2bits(j % 2 ^ (i * 8)) % 8 ≠ 0, 1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), i→n(i) < # int2bits(j % 2 ^ (i * 8)) / 8 + 1, 0 < i, j + 2 ^ (i * 8 - 1) < 0, j + 2 ^ (i * 8) ≤ 0, j + 2 ^ (i * 8) < i, j % 2 ^ (i * 8) < 0, 0 ≤ i ⊦ int2bits(j % 2 ^ (i * 8)).rest = [], j % 2 ^ (i * 8) = 0, bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = lastn(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 177:
1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), 0 ≤ i, j % 2 ^ (i * 8) < 0, j + 2 ^ (i * 8) < i, j + 2 ^ (i * 8) ≤ 0, j + 2 ^ (i * 8 - 1) < 0, 0 < i ⊦ bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = lastn(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest, j % 2 ^ (i * 8) = 0, int2bits(j % 2 ^ (i * 8)).rest = []

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ i;% ⊦ 0 < j → abs(j) = j;% ⊦ i→n(i) ≤ i→n(j) ↔ abs(i) ≤ abs(j);% ⊦ 0 + m = m;%n ≠ 0 ⊦ (n0 + m * n) / n = n0 / n + m;%n ≠ 0 ⊦ (m + n0 * n) % n = m % n;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ i * j = j * i;