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Proof step 16 from proof int2bytes-le255 (previous step)

0 ≤ 7 ∧ 2 ^ 7 ≤ i ∧ i < 2 ^ (7 + 1), 8 < i→n(7 + 2), 127 < i, i ≤ 255, 0 < i, 0 ≤ i ⊦ bits2bytes(firstn(i→n(7 + 2) - 8, uint2bits(i))) + i2b(bits2int(lastn(8, uint2bits(i)))) ' = i2b(0) ' + i2b(i - 256) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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