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Proof step 108 from proof int2bytes-rec-neg-02 (previous step)

# int2bits(j % 2 ^ (i * 8)) % 8 ≠ 0, 1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), i→n(i) + 1 = # int2bits(j % 2 ^ (i * 8)) / 8 + 1, 8 * i→n(i) ≤ 7, j + 2 ^ (i * 8) < 0, j % 2 ^ (i * 8) < 0, 0 < i ⊦ i = 0, int2bits(j % 2 ^ (i * 8)).rest = [], j % 2 ^ (i * 8) = 0, bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = bits2bytes(int2bits(j % 2 ^ (i * 8))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 109:
1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), 0 < i, j % 2 ^ (i * 8) < 0, j + 2 ^ (i * 8) < 0, 8 * i→n(i) ≤ 7 ⊦ bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = bits2bytes(int2bits(j % 2 ^ (i * 8))).rest, j % 2 ^ (i * 8) = 0, int2bits(j % 2 ^ (i * 8)).rest = [], i = 0

Rule argument

Additional information

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