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Proof step 86 from proof int2bytes-rec-neg-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 87:
1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), 0 ≤ i, j % 2 ^ (i * 8) < 0, j + 2 ^ (i * 8) < 0, 0 < i ⊦ bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = fix(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 + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;%n ≠ 0 ⊦ (m * n) % n = 0;% ⊦ i * j = j * i;