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Proof step 128 from proof int2bytes-rec-neg-02 (previous step)

¬(# bits2bytes(int2bits(j % 2 ^ (i * 8))) < i→n(i) + 1 ∧ bytes2int(bits2bytes(int2bits(j % 2 ^ (i * 8)))) < 0), # int2bits(j % 2 ^ (i * 8)) = 8 * i→n(i), 0 < i, j + 2 ^ (i * 8) < 0, j % 2 ^ (i * 8) < 0, 0 ≤ i ⊦ j % 2 ^ (i * 8) = 0, bits2bytes(int2bits(j % 2 ^ (i * 8))) = fix(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ i * j = j * i;