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Proof step 16 from proof bits2bytes-rec-02 (previous step)

8 ≠ 0, # bits = n0 * 8 + 8, n0 ≤ 1, ∀ bits0. # bits0 < # bits → true → # bits0 % 8 = 0 ∧ 8 < # bits0 → bits2bytes(bits0) = bits2bytes(firstn(8, bits0)) + bits2bytes(restn(8, bits0)) ⊦ bits2bytes(firstn(n0 * 8, bits)) + i2b(bits2int(lastn(8, bits))) ' = bits2bytes(firstn(8, bits)) + bits2bytes(restn(8, bits)), bits = [], n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# bits = 16, ∀ bits0. # bits0 < 16 → true → # bits0 % 8 = 0 ∧ 8 < # bits0 → bits2bytes(bits0) = bits2bytes(firstn(8, bits0)) + bits2bytes(restn(8, bits0)) ⊦ bits = [], i2b(bits2int(lastn(8, bits))) ' = bits2bytes(restn(8, bits))

Rule argument

Additional information

Used simplifier rules: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;% ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1);