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Proof step 7 from proof bits2bytes-rec-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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