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Proof step 34 from proof bits2bytes-rec-02 (previous step)

8 + 8 ≤ # bits, # bits = 16 + n0, n0 % 8 = 0, ∀ bits0. # bits0 < (n0 + 8) + 8 → true → # bits0 % 8 = 0 ∧ 8 < # bits0 → bits2bytes(bits0) = bits2bytes(firstn(8, bits0)) + bits2bytes(restn(8, bits0)) ⊦ n0 = 0, bits = [], bits2bytes(restn(8, firstn(n0 + 8, bits))) + i2b(bits2int(lastn(8, bits))) ' = bits2bytes(firstn(n0, restn(8, bits))) + i2b(bits2int(lastn(8, bits))) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m + n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ (* n) ≠ m0 + (* n0);%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes1 + bytes0 ↔ bytes = bytes1;