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Proof step 121 from proof int2bytes-rec-pos-any (previous step)

# bits2bytes(int2bits(j % 2 ^ (i * 8))) < i→n(i) + 1, i→n(i * 8) % 8 = 0, # int2bits(j % 2 ^ (i * 8)) = i→n(i * 8), 0 ≤ j % 2 ^ (i * 8), 0 < i, 2 ^ (i * 8) ≤ j ⊦ 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

Proof step 122:
i→n(i * 8) % 8 = 0, # int2bits(j % 2 ^ (i * 8)) = i→n(i * 8), 0 ≤ j % 2 ^ (i * 8), 0 < i, 2 ^ (i * 8) ≤ j, # bits2bytes(int2bits(j % 2 ^ (i * 8))) ≤ i→n(i) ⊦ bits2bytes(int2bits(j % 2 ^ (i * 8))) = fix(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest

Rule argument

Additional information

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