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Proof step 142 from proof int2bytes-rec-pos-any (previous step)

# int2bits(j % 2 ^ (i * 8)) % 8 = 0 ∧ int2bits(j % 2 ^ (i * 8)) ≠ [], # int2bits(j % 2 ^ (i * 8)) = i→n(i * 8), i→n(i * 8) % 8 = 0, i→n(i) < # int2bits(j % 2 ^ (i * 8)) / 8, 2 ^ (i * 8) ≤ j, 0 < i, 0 ≤ j % 2 ^ (i * 8) ⊦ 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 143:
# int2bits(j % 2 ^ (i * 8)) = i→n(i * 8), i→n(i * 8) % 8 = 0, 0 ≤ j % 2 ^ (i * 8), 0 < i, 2 ^ (i * 8) ≤ j, i→n(i) < i→n(i * 8) / 8 ⊦ 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 ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ int2bits(j) ≠ [];% ⊦ i * j = j * i;