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Proof step 29 from proof bits2bytes-rec-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
bits2bytes(firstn(n0 + 8, bits)) = bits2bytes(firstn(8, bits)) + bits2bytes(restn(8, firstn(n0 + 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(firstn(n0 + 8, bits)) + i2b(bits2int(lastn(8, bits))) ' = bits2bytes(firstn(8, bits)) + bits2bytes(firstn(n0, restn(8, bits))) + i2b(bits2int(lastn(8, restn(8, bits)))) '

Rule argument

Additional information

Used simplifier rules: n < n0 ⊦ m + (* n) ≤ m0 + (* n0) ↔ m ≤ m0 + (n0 - n);% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m ≤ m + n;% ⊦ m + n = n + m;