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Proof step 30 from proof int2bytes-rec-pos (previous step)

# uint2bits(i % 256) ≤ 9, uint2bits(i) = uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest, 8 < (9 - 1) + # uint2bits(i / 256), 255 < i, 0 < i, 0 ≤ i ⊦ bits2bytes(firstn((9 - 1) + # uint2bits(i / 256) - 8, uint2bits(i))) + i2b(bits2int(lastn(8, uint2bits(i)))) ' = bits2bytes(uint2bits(i / 256)) + bits2bytes(int2bits(i % 256)).last ', fillsgn(9, uint2bits(i % 256)) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i / j;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < m + n ↔ n ≠ 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;