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Proof step 21 from proof int2bytes-rec-neg-04 (previous step)

i→n(i * 8) % 8 = 0, 0 < i, j0 * 2 ^ (i * 8) + 2 ^ (i * 8 - 1) < 0 ⊦ bits2bytes(int2bits(j0)) + bits2bytes(mklist(false, i→n(i * 8))) = bits2bytes(int2bits(j0)) + fix(i→n(i) + 1, i2b(0) ').rest, j0 = 0, i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
i→n(i * 8) % 8 = 0, 0 < i, j0 * 2 ^ (i * 8) + 2 ^ (i * 8 - 1) < 0 ⊦ j0 = 0, i = 0, bits2bytes(mklist(false, i→n(i * 8))) = fix(i→n(i) + 1, i2b(0) ').rest

Rule argument

Additional information

Used simplifier rules: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;