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Proof step 184 from proof int2bytes-rec-neg-05 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 185:
1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), i→n(i) + 1 = # bits2bytes(int2bits(j % 2 ^ (i * 8))), 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, 8 * i→n(i) ≤ 7 ⊦ 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

Rule argument

Additional information

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